stablegradients/pie-gem5-pairs huggingface dataset:
// Slower C++ solution:
#include"stdio.h"

#include"stdlib.h"

#include"string.h"

#include"algorithm"

#include"queue"

using namespace std;

struct mmm

{

	int x,y;

}vv;

int dir[4][2] = {0,-1,-1,0,0,1,1,0};

char xx[1005][1005];

int bb[11][2];

queue <mmm> qq;

int vis[1005][1005];

int bfs(char t)

{

	int sum = 0;

	int pl = 0;

	while(!qq.empty())

	{

		vv = qq.front();

		int x0 = vv.x;

		int y0 = vv.y;

		//printf("%d %d %d\n",x0,y0,vis[x0][y0]);

		qq.pop();

		if(xx[x0][y0] == t && !pl)

		{

			sum = vis[x0][y0];

			pl = 1;

		}

		for(int i = 0;i < 4;i++)

		{

			int xt = x0 + dir[i][0];

			int yt = y0 + dir[i][1];

			if(xx[xt][yt] != '#' && xx[xt][yt] != 'X' && vis[xt][yt] == -1)

			{

				vis[xt][yt] = vis[x0][y0] + 1;

				vv.x = xt;

				vv.y = yt;

				qq.push(vv);

			}

		}

	}

	return sum;

}

int main()

{

	int a,b,c,d,x1,y1,sum;

	while(~scanf("%d %d %d",&a,&b,&c))

	{

	//	if(a == 0 && b == 0)

	//	break;

		memset(xx,'#',sizeof(xx));

		getchar();

		for(int i = 1;i <= a;i++)

		{

			for(int j = 1;j <= b;j++)

			{

				xx[i][j] = getchar();

				if(xx[i][j] == 'S')

				{

					x1 = i;

					y1 = j;

				}

				else if(xx[i][j] != 'X')

				{

					bb[xx[i][j] - '0'][0] = i;

					bb[xx[i][j] - '0'][1] = j;

				}

			}

			getchar();

		}

		bb[0][0] = x1;

		bb[0][1] = y1;

		sum = 0;

		for(int i = 1;i <= c;i++)

		{

			vv.x = bb[i-1][0];

			vv.y = bb[i-1][1];

			memset(vis,-1,sizeof(vis));

			vis[vv.x][vv.y] = 0;

			qq.push(vv);

			sum += bfs(i + '0');

		}

		printf("%d\n",sum);

	}

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.212218722293073x):
#include<stdio.h>

#include<string.h>

#include<queue>

using namespace std;

const int maxn=1005;

int n,m,k,dx[]={-1,1,0,0},dy[]={0,0,1,-1},vis[maxn][maxn];

char map[maxn][maxn];

struct maze

{

    maze(int xx=0,int yy=0,int tt=0):x(xx),y(yy),time(tt){}

	int x,y,time;

};

maze bfs(int bx,int by,char ch)

{

	memset(vis,0,sizeof(vis));

	queue<maze> q;

	maze st(bx,by,0);

	q.push(st);vis[bx][by]=1;

	while(!q.empty())

	{

		st=q.front();q.pop();

		if(map[st.x][st.y]==ch)

		{

			return st;

		}

		for(int i=0;i<4;++i)

		{

			int tx=st.x+dx[i],ty=st.y+dy[i];

			if(tx<0||tx>=n||ty<0||ty>=m||map[tx][ty]=='X')//越界和?

			{

				continue;

			}

			maze tp(tx,ty,st.time+1);

			if(!vis[tx][ty])

			{

				vis[tx][ty]=1;

				q.push(tp);

			}

		}

	}

}

int main()

{



	while(~scanf("%d%d%d",&n,&m,&k))

	{

		int bx,by;

		for(int i=0;i<n;++i)

		{

			scanf("%s",map[i]);

			for(int j=0;j<m;++j)

			{

				if(map[i][j]=='S')

				{

					bx=i;by=j;

					

					map[i][j]='0'; //第0个起点

				}

			}

		}

		int ans=0;

		for(int i=0;i<k;++i)

		{

			maze tp=bfs(bx,by,map[bx][by]+1);

			ans+=tp.time;

			bx=tp.x;by=tp.y;

		}

		printf("%d\n",ans);

	}

	return 0;

}
// Slower C++ solution:
/*

*

*/

#include<iostream>

#include<cstring>

#include<cstdio>

#include<cmath>

#include<string>

#include<queue>

#define CLR(ARR,NUM) memset(ARR,NUM,sizeof(ARR))

#define GETNUM(NUM) scanf("%d",&NUM)

#define faster_io() ios_base::sync_with_stdio(false)

using namespace std;

const int MAXN = 1010;

char map[MAXN][MAXN];

bool vis[10][MAXN][MAXN];

struct node{

	int x;

	int y;

	int deep;

	int n;

};

int dx[]={0,0,1,-1};

int dy[]={1,-1,0,0};



int main()

{

	CLR(vis,false);

	CLR(map,'X');

	int h,w,n;

	cin >> h >> w >> n;

	getchar();

	int sx,sy;

	for(int i=1;i<=h;i++){

		for(int j=1;j<=w;j++){

			map[i][j]=getchar();

			if(map[i][j]=='S'){

				sx=i;

				sy=j;

			}

		}

		getchar();

	}

	node t;

	t.x=sx;

	t.y=sy;

	t.deep=0;

	t.n=0;

	queue<node>q;

	vis[0][sx][sy]=true;

	q.push(t);

	int ans = -1;

	while(!q.empty()){

		node cur=q.front();

		if(cur.n==n){

			ans=cur.deep;

			break;

		}

		q.pop();

		for(int i=0;i<4;i++){

			int tx=cur.x+dx[i];

			int ty=cur.y+dy[i];

			if(map[tx][ty]!='X'&&!vis[cur.n][tx][ty]){

				node t = cur;

				t.x=tx;

				t.y=ty;

				t.deep++;

				if(map[tx][ty]=='1'+t.n){

					t.n++;

				}

				vis[t.n][tx][ty]=true;

				q.push(t);

			}

		}

	}

	cout << ans << endl;

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 2.549699565742426x):
#include<stdio.h>

#include<string.h>

#include<queue>

using namespace std;

const int maxn=1005;

int n,m,k,dx[]={-1,1,0,0},dy[]={0,0,1,-1},vis[maxn][maxn];

char map[maxn][maxn];

struct maze

{

    maze(int xx=0,int yy=0,int tt=0):x(xx),y(yy),time(tt){}

	int x,y,time;

};

maze bfs(int bx,int by,char ch)

{

	memset(vis,0,sizeof(vis));

	queue<maze> q;

	maze st(bx,by,0);

	q.push(st);vis[bx][by]=1;

	while(!q.empty())

	{

		st=q.front();q.pop();

		if(map[st.x][st.y]==ch)

		{

			return st;

		}

		for(int i=0;i<4;++i)

		{

			int tx=st.x+dx[i],ty=st.y+dy[i];

			if(tx<0||tx>=n||ty<0||ty>=m||map[tx][ty]=='X')//越界和?

			{

				continue;

			}

			maze tp(tx,ty,st.time+1);

			if(!vis[tx][ty])

			{

				vis[tx][ty]=1;

				q.push(tp);

			}

		}

	}

}

int main()

{



	while(~scanf("%d%d%d",&n,&m,&k))

	{

		int bx,by;

		for(int i=0;i<n;++i)

		{

			scanf("%s",map[i]);

			for(int j=0;j<m;++j)

			{

				if(map[i][j]=='S')

				{

					bx=i;by=j;

					

					map[i][j]='0'; //第0个起点

				}

			}

		}

		int ans=0;

		for(int i=0;i<k;++i)

		{

			maze tp=bfs(bx,by,map[bx][by]+1);

			ans+=tp.time;

			bx=tp.x;by=tp.y;

		}

		printf("%d\n",ans);

	}

	return 0;

}
// Slower C++ solution:
#include <queue>

#include <cstring>

#include <iostream>

using namespace std;



const int N = 1111;

const int cx[] = {0, 0, 1, -1}, cy[] = {1, -1, 0, 0};

char maze[N][N];

int step[N][N];

int h, w, n, sx, sy, ans;



int bfs(int cheese){

    memset(step, -1, sizeof(step));

    queue<pair<int, int> > que;

    que.push(make_pair(sx, sy));

    step[sx][sy] = 0;

    while(!que.empty()){

        int x = que.front().first, y = que.front().second;

        que.pop();

        for(int i = 0; i < 4; ++i){

            int nx = x + cx[i], ny = y + cy[i];

            if(nx < 0 || h <= nx || ny < 0 || w <= ny || maze[nx][ny] == 'X' || step[nx][ny] != -1) continue;

            step[nx][ny] = step[x][y] + 1;

            if(maze[nx][ny] - '0' == cheese){

                sx = nx;

                sy = ny;

                return step[nx][ny];

            }

            que.push(make_pair(nx, ny));

        }

    }

    return 0;

}



int main(){

    ios::sync_with_stdio(false);

    cin.tie(NULL);

    while(cin >> h >> w >> n){

        for(int i = 0; i < h; ++i)

            for(int j = 0; j < w; ++j){

                cin >> maze[i][j];

                if(maze[i][j] == 'S'){

                    sx = i;

                    sy = j;

                }

            }

        ans = 0;

        for(int i = 1; i <= n; ++i)

            ans += bfs(i);

        cout << ans << endl;

    }

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.2116307585446213x):
#include<stdio.h>

#include<string.h>

#include<queue>

using namespace std;

const int maxn=1005;

int n,m,k,dx[]={-1,1,0,0},dy[]={0,0,1,-1},vis[maxn][maxn];

char map[maxn][maxn];

struct maze

{

    maze(int xx=0,int yy=0,int tt=0):x(xx),y(yy),time(tt){}

	int x,y,time;

};

maze bfs(int bx,int by,char ch)

{

	memset(vis,0,sizeof(vis));

	queue<maze> q;

	maze st(bx,by,0);

	q.push(st);vis[bx][by]=1;

	while(!q.empty())

	{

		st=q.front();q.pop();

		if(map[st.x][st.y]==ch)

		{

			return st;

		}

		for(int i=0;i<4;++i)

		{

			int tx=st.x+dx[i],ty=st.y+dy[i];

			if(tx<0||tx>=n||ty<0||ty>=m||map[tx][ty]=='X')//越界和?

			{

				continue;

			}

			maze tp(tx,ty,st.time+1);

			if(!vis[tx][ty])

			{

				vis[tx][ty]=1;

				q.push(tp);

			}

		}

	}

}

int main()

{



	while(~scanf("%d%d%d",&n,&m,&k))

	{

		int bx,by;

		for(int i=0;i<n;++i)

		{

			scanf("%s",map[i]);

			for(int j=0;j<m;++j)

			{

				if(map[i][j]=='S')

				{

					bx=i;by=j;

					

					map[i][j]='0'; //第0个起点

				}

			}

		}

		int ans=0;

		for(int i=0;i<k;++i)

		{

			maze tp=bfs(bx,by,map[bx][by]+1);

			ans+=tp.time;

			bx=tp.x;by=tp.y;

		}

		printf("%d\n",ans);

	}

	return 0;

}
// Slower C++ solution:


#include <iostream>

#include <cstdio>

#include <cstring>

#include <queue>



using namespace std;

const int maxn = 1e3 + 5;

const int dir[4][2] = {{1,0},{0,1},{-1,0},{0,-1}};

int mp[maxn][maxn];

//bool vis[maxn][maxn];

int d[maxn][maxn];

struct point{

    int x,y;

};

int h,w,n;

 int dis = 0;

void bfs(point &p,int cnt)

{

   // memset(vis, 0, sizeof(vis));

    memset(d, 0, sizeof(d));

    queue<point> q;

    q.push(p);

    while (!q.empty()) {

        point t = q.front();q.pop();

        //vis[t.x][t.y] = true;

        if(mp[t.x][t.y] == cnt + 1) {dis += d[t.x][t.y];p.x = t.x,p.y = t.y;return ;}

        for (int i = 0; i < 4; i ++) {

            int tx = t.x + dir[i][0];

            int ty = t.y + dir[i][1];

            if (tx >= 0 && tx < h && ty >= 0 && ty < w && !d[tx][ty] && mp[tx][ty] != -1) {

                q.push({tx,ty});

                d[tx][ty] = 1 + d[t.x][t.y];

            }

        }

    }

}

int main()

{

    while (scanf("%d%d%d",&h,&w,&n) != EOF) {

        getchar();

        char ch;

        memset(mp, -1, sizeof(mp));

       // int sx,sy;

        point s;

        for (int i = 0; i < h; i ++) {

            for (int j = 0; j < w; j ++) {

                scanf("%c",&ch);

                if(ch == 'S') {s.x = i,s.y = j;mp[i][j] = 0;}

                else if(ch == '.') {mp[i][j] = 0;}

                else if(isdigit(ch)) {mp[i][j] = ch - '0';}

            }

            getchar();

        }

       

        dis = 0;

        for (int i = 0; i < n; i ++) {

            bfs(s,i);

        }

        printf("%d\n",dis);

    }

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.86643085031677x):
#include<stdio.h>

#include<string.h>

#include<queue>

using namespace std;

const int maxn=1005;

int n,m,k,dx[]={-1,1,0,0},dy[]={0,0,1,-1},vis[maxn][maxn];

char map[maxn][maxn];

struct maze

{

    maze(int xx=0,int yy=0,int tt=0):x(xx),y(yy),time(tt){}

	int x,y,time;

};

maze bfs(int bx,int by,char ch)

{

	memset(vis,0,sizeof(vis));

	queue<maze> q;

	maze st(bx,by,0);

	q.push(st);vis[bx][by]=1;

	while(!q.empty())

	{

		st=q.front();q.pop();

		if(map[st.x][st.y]==ch)

		{

			return st;

		}

		for(int i=0;i<4;++i)

		{

			int tx=st.x+dx[i],ty=st.y+dy[i];

			if(tx<0||tx>=n||ty<0||ty>=m||map[tx][ty]=='X')//越界和?

			{

				continue;

			}

			maze tp(tx,ty,st.time+1);

			if(!vis[tx][ty])

			{

				vis[tx][ty]=1;

				q.push(tp);

			}

		}

	}

}

int main()

{



	while(~scanf("%d%d%d",&n,&m,&k))

	{

		int bx,by;

		for(int i=0;i<n;++i)

		{

			scanf("%s",map[i]);

			for(int j=0;j<m;++j)

			{

				if(map[i][j]=='S')

				{

					bx=i;by=j;

					

					map[i][j]='0'; //第0个起点

				}

			}

		}

		int ans=0;

		for(int i=0;i<k;++i)

		{

			maze tp=bfs(bx,by,map[bx][by]+1);

			ans+=tp.time;

			bx=tp.x;by=tp.y;

		}

		printf("%d\n",ans);

	}

	return 0;

}
// Slower C++ solution:
/*

*

*/

#include<iostream>

#include<cstring>

#include<cstdio>

#include<cmath>

#include<string>

#include<queue>

#define CLR(ARR,NUM) memset(ARR,NUM,sizeof(ARR))

#define GETNUM(NUM) scanf("%d",&NUM)

#define faster_io() ios_base::sync_with_stdio(false)

using namespace std;

const int MAXN = 1010;

char map[MAXN][MAXN];

bool vis[10][MAXN][MAXN];

struct node{

	int x;

	int y;

	int deep;

	int n;

};

int dx[]={0,0,1,-1};

int dy[]={1,-1,0,0};



int main()

{

	CLR(vis,false);

	CLR(map,'X');

	int h,w,n;

	cin >> h >> w >> n;

	getchar();

	int sx,sy;

	for(int i=1;i<=h;i++){

		for(int j=1;j<=w;j++){

			map[i][j]=getchar();

			if(map[i][j]=='S'){

				sx=i;

				sy=j;

			}

		}

		getchar();

	}

	node t;

	t.x=sx;

	t.y=sy;

	t.deep=0;

	t.n=0;

	queue<node>q;

	vis[0][sx][sy]=true;

	q.push(t);

	int ans = -1;

	while(!q.empty()){

		node cur=q.front();

		if(cur.n==n){

			ans=cur.deep;

			break;

		}

		q.pop();

		for(int i=0;i<4;i++){

			int tx=cur.x+dx[i];

			int ty=cur.y+dy[i];

			if(map[tx][ty]!='X'&&!vis[cur.n][tx][ty]){

				node t = cur;

				t.x=tx;

				t.y=ty;

				t.deep++;

				if(map[tx][ty]=='1'+t.n){

					t.n++;

				}

				vis[t.n][tx][ty]=true;

				q.push(t);

			}

		}

	}

	cout << ans << endl;

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 12.725204817520183x):
#include<iostream>

#include<algorithm>

#include<queue>

const int INF = 100000000;

using namespace std;

typedef pair<int, int> P;

char a[1005][1005];

int N, M;

int sx, sy;

int gx, gy;

int d[1005][1005];

int dx[4] = { 1,0,-1,0 };

int dy[4] = { 0,1,0,-1 };

int bfs()

{

	queue<P> que;

	for (int i = 0; i < N; i++)

		for (int j = 0; j < M; j++)

			d[i][j] = INF;

	que.push(P(sx, sy));

	d[sx][sy] = 0;

	while (que.size())

	{

		P p = que.front();

		que.pop();

		if (p.first == gx&&p.second == gy)

			break;

		for (int i = 0; i < 4; i++)

		{

			int nx = p.first + dx[i];

			int ny = p.second + dy[i];

			if (nx >= 0 && nx < N&&ny >= 0 && ny < M&&a[nx][ny] != 'X'&&d[nx][ny] == INF)

			{

				que.push(P(nx, ny));

				d[nx][ny] = d[p.first][p.second] + 1;

			}

		}



	}

	return d[gx][gy];

}

int main(void)

{

	while (cin >> N >> M)

	{

		int n, sum = 0;

		cin >> n;

		for (int i = 0; i < N; i++)

			for (int j = 0; j < M; j++)

				cin >> a[i][j];

		for (int j = 0; j < N; j++)

			for (int k = 0; k < M; k++)

				if (a[j][k] == 'S')

				{

					sx = j;

					sy = k;

				}

		for (int j = 0; j < N; j++)

			for (int k = 0; k < M; k++)

				if (a[j][k] == '1')

				{

					gx = j;

					gy = k;

				}

		sum += bfs();

		for (int i = 0; i < n - 1; i++)

		{

			for (int j = 0; j < N; j++)

				for (int k = 0; k < M; k++)

				{

					if (a[j][k] == (char)(i+49))

					{

						sx = j;

						sy = k;

					}

					if (a[j][k] == (char)(i+50))

					{

						gx = j;

						gy = k;

					}

				}

			sum += bfs();



		}

		cout << sum << endl;

	}

}
// Slower C++ solution:
#include <queue>

#include <cstring>

#include <iostream>

using namespace std;



const int N = 1111;

const int cx[] = {0, 0, 1, -1}, cy[] = {1, -1, 0, 0};

char maze[N][N];

int step[N][N];

int h, w, n, sx, sy, ans;



int bfs(int cheese){

    memset(step, -1, sizeof(step));

    queue<pair<int, int> > que;

    que.push(make_pair(sx, sy));

    step[sx][sy] = 0;

    while(!que.empty()){

        int x = que.front().first, y = que.front().second;

        que.pop();

        for(int i = 0; i < 4; ++i){

            int nx = x + cx[i], ny = y + cy[i];

            if(nx < 0 || h <= nx || ny < 0 || w <= ny || maze[nx][ny] == 'X' || step[nx][ny] != -1) continue;

            step[nx][ny] = step[x][y] + 1;

            if(maze[nx][ny] - '0' == cheese){

                sx = nx;

                sy = ny;

                return step[nx][ny];

            }

            que.push(make_pair(nx, ny));

        }

    }

    return 0;

}



int main(){

    ios::sync_with_stdio(false);

    cin.tie(NULL);

    while(cin >> h >> w >> n){

        for(int i = 0; i < h; ++i)

            for(int j = 0; j < w; ++j){

                cin >> maze[i][j];

                if(maze[i][j] == 'S'){

                    sx = i;

                    sy = j;

                }

            }

        ans = 0;

        for(int i = 1; i <= n; ++i)

            ans += bfs(i);

        cout << ans << endl;

    }

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 6.047084830246709x):
#include<iostream>

#include<algorithm>

#include<queue>

const int INF = 100000000;

using namespace std;

typedef pair<int, int> P;

char a[1005][1005];

int N, M;

int sx, sy;

int gx, gy;

int d[1005][1005];

int dx[4] = { 1,0,-1,0 };

int dy[4] = { 0,1,0,-1 };

int bfs()

{

	queue<P> que;

	for (int i = 0; i < N; i++)

		for (int j = 0; j < M; j++)

			d[i][j] = INF;

	que.push(P(sx, sy));

	d[sx][sy] = 0;

	while (que.size())

	{

		P p = que.front();

		que.pop();

		if (p.first == gx&&p.second == gy)

			break;

		for (int i = 0; i < 4; i++)

		{

			int nx = p.first + dx[i];

			int ny = p.second + dy[i];

			if (nx >= 0 && nx < N&&ny >= 0 && ny < M&&a[nx][ny] != 'X'&&d[nx][ny] == INF)

			{

				que.push(P(nx, ny));

				d[nx][ny] = d[p.first][p.second] + 1;

			}

		}



	}

	return d[gx][gy];

}

int main(void)

{

	while (cin >> N >> M)

	{

		int n, sum = 0;

		cin >> n;

		for (int i = 0; i < N; i++)

			for (int j = 0; j < M; j++)

				cin >> a[i][j];

		for (int j = 0; j < N; j++)

			for (int k = 0; k < M; k++)

				if (a[j][k] == 'S')

				{

					sx = j;

					sy = k;

				}

		for (int j = 0; j < N; j++)

			for (int k = 0; k < M; k++)

				if (a[j][k] == '1')

				{

					gx = j;

					gy = k;

				}

		sum += bfs();

		for (int i = 0; i < n - 1; i++)

		{

			for (int j = 0; j < N; j++)

				for (int k = 0; k < M; k++)

				{

					if (a[j][k] == (char)(i+49))

					{

						sx = j;

						sy = k;

					}

					if (a[j][k] == (char)(i+50))

					{

						gx = j;

						gy = k;

					}

				}

			sum += bfs();



		}

		cout << sum << endl;

	}

}
// Slower C++ solution:
#include<iostream>

#include<cstdio>

#include<queue>

#include<utility>

#include<cstring>

using namespace std;

char map[1005][1005];

bool vis[1005][1005];

int dx[]={0,0,-1,1};

int dy[]={1,-1,0,0};

int sx,sy;

int h,w;

int bfs(int n){

	memset(vis,false,sizeof(vis));

	queue<pair<pair<int,int>,int> >que;

	que.push(make_pair(make_pair(sx,sy),0));

	while(!que.empty()){

		pair<int,int>now=que.front().first;

		int cnt=que.front().second;

		que.pop();

		if(map[now.first][now.second]==n+48){

			sx=now.first;

			sy=now.second;

			return cnt;

		}

		for(int i=0;i<4;i++){

			int _x=now.first+dx[i];

			int _y=now.second+dy[i];

			if(_x<0||_y<0||_x>=h||_y>=w||map[_x][_y]=='X'||vis[_x][_y]){

				continue;

			}

			vis[_x][_y]=true;

			que.push(make_pair(make_pair(_x,_y),cnt+1));

		}

	}

}

int main(){

	int n;

	while(cin>>h>>w>>n){

		memset(map,0,sizeof(map));

		for(int i=0;i<h;i++){

			scanf("%s",map[i]);

		}

		for(int i=0;i<h;i++){

			for(int j=0;j<w;j++){

				if(map[i][j]=='S'){

					sx=i;

					sy=j;

					map[i][j]=='.';

					break;

				}

			}

		}

		int count=0;

		for(int k=1;k<=n;k++){

			count+=bfs(k);

		}

		cout<<count<<endl;

	}

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 3.007098462653392x):
#include<iostream>

#include<algorithm>

#include<queue>

const int INF = 100000000;

using namespace std;

typedef pair<int, int> P;

char a[1005][1005];

int N, M;

int sx, sy;

int gx, gy;

int d[1005][1005];

int dx[4] = { 1,0,-1,0 };

int dy[4] = { 0,1,0,-1 };

int bfs()

{

	queue<P> que;

	for (int i = 0; i < N; i++)

		for (int j = 0; j < M; j++)

			d[i][j] = INF;

	que.push(P(sx, sy));

	d[sx][sy] = 0;

	while (que.size())

	{

		P p = que.front();

		que.pop();

		if (p.first == gx&&p.second == gy)

			break;

		for (int i = 0; i < 4; i++)

		{

			int nx = p.first + dx[i];

			int ny = p.second + dy[i];

			if (nx >= 0 && nx < N&&ny >= 0 && ny < M&&a[nx][ny] != 'X'&&d[nx][ny] == INF)

			{

				que.push(P(nx, ny));

				d[nx][ny] = d[p.first][p.second] + 1;

			}

		}



	}

	return d[gx][gy];

}

int main(void)

{

	while (cin >> N >> M)

	{

		int n, sum = 0;

		cin >> n;

		for (int i = 0; i < N; i++)

			for (int j = 0; j < M; j++)

				cin >> a[i][j];

		for (int j = 0; j < N; j++)

			for (int k = 0; k < M; k++)

				if (a[j][k] == 'S')

				{

					sx = j;

					sy = k;

				}

		for (int j = 0; j < N; j++)

			for (int k = 0; k < M; k++)

				if (a[j][k] == '1')

				{

					gx = j;

					gy = k;

				}

		sum += bfs();

		for (int i = 0; i < n - 1; i++)

		{

			for (int j = 0; j < N; j++)

				for (int k = 0; k < M; k++)

				{

					if (a[j][k] == (char)(i+49))

					{

						sx = j;

						sy = k;

					}

					if (a[j][k] == (char)(i+50))

					{

						gx = j;

						gy = k;

					}

				}

			sum += bfs();



		}

		cout << sum << endl;

	}

}
// Slower C++ solution:
#include<stdio.h>

#include<string.h>

#include<queue>

#include<algorithm>

#define MAX_N 1002

using namespace std;

struct node{int x,y;};

char _map[MAX_N][MAX_N];

int dp[MAX_N][MAX_N];

int ox[]={0,0,1,-1};

int oy[]={1,-1,0,0};

int S_x,S_y;

int h,w,cnt;



node make_node(int x,int y){

	node t;

	t.x=x,t.y=y;

	return t;

}

void solve(){

	queue<node> que;

	int step=0;

	

	for(int i=1;i<=cnt;i++){

		que.push(make_node(S_x,S_y));

		_map[S_x][S_y]='.';

		memset(dp,-1,sizeof(dp));

		dp[S_x][S_y]=step;

		while(!que.empty()){

			node t=que.front();que.pop();

			if(_map[t.x][t.y]=='0'+i){

				step=dp[t.x][t.y];

				S_x=t.x,S_y=t.y;

				break;

			}

			for(int j=0;j<4;j++){

				int x=t.x+ox[j];

				int y=t.y+oy[j];

				if(0<=x&&x<h&&0<=y&&y<w&&_map[x][y]!='X'&&dp[x][y]==-1){

					que.push(make_node(x,y));

					dp[x][y]=dp[t.x][t.y]+1;

				}

			}

		}

		while(!que.empty()) que.pop();

	}

	

	printf("%d\n",step);

}



int main()

{

	while(scanf("%d%d%d",&h,&w,&cnt)!=EOF){

		for(int i=0;i<h;i++)

			scanf("%s",_map[i]);

		for(int j=0;j<h;j++)

			for(int k=0;k<w;k++)

				if(_map[j][k]=='S')

					S_x=j,S_y=k,j=h,k=w;

		solve();

	}

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 4.965835299758284x):
#include<iostream>

#include<algorithm>

#include<queue>

const int INF = 100000000;

using namespace std;

typedef pair<int, int> P;

char a[1005][1005];

int N, M;

int sx, sy;

int gx, gy;

int d[1005][1005];

int dx[4] = { 1,0,-1,0 };

int dy[4] = { 0,1,0,-1 };

int bfs()

{

	queue<P> que;

	for (int i = 0; i < N; i++)

		for (int j = 0; j < M; j++)

			d[i][j] = INF;

	que.push(P(sx, sy));

	d[sx][sy] = 0;

	while (que.size())

	{

		P p = que.front();

		que.pop();

		if (p.first == gx&&p.second == gy)

			break;

		for (int i = 0; i < 4; i++)

		{

			int nx = p.first + dx[i];

			int ny = p.second + dy[i];

			if (nx >= 0 && nx < N&&ny >= 0 && ny < M&&a[nx][ny] != 'X'&&d[nx][ny] == INF)

			{

				que.push(P(nx, ny));

				d[nx][ny] = d[p.first][p.second] + 1;

			}

		}



	}

	return d[gx][gy];

}

int main(void)

{

	while (cin >> N >> M)

	{

		int n, sum = 0;

		cin >> n;

		for (int i = 0; i < N; i++)

			for (int j = 0; j < M; j++)

				cin >> a[i][j];

		for (int j = 0; j < N; j++)

			for (int k = 0; k < M; k++)

				if (a[j][k] == 'S')

				{

					sx = j;

					sy = k;

				}

		for (int j = 0; j < N; j++)

			for (int k = 0; k < M; k++)

				if (a[j][k] == '1')

				{

					gx = j;

					gy = k;

				}

		sum += bfs();

		for (int i = 0; i < n - 1; i++)

		{

			for (int j = 0; j < N; j++)

				for (int k = 0; k < M; k++)

				{

					if (a[j][k] == (char)(i+49))

					{

						sx = j;

						sy = k;

					}

					if (a[j][k] == (char)(i+50))

					{

						gx = j;

						gy = k;

					}

				}

			sum += bfs();



		}

		cout << sum << endl;

	}

}