stablegradients/pie-gem5-pairs huggingface dataset:
// 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: 2.203059201171553x):
#include<cstdio>

#include<cstdlib>

#include<cmath>

#include<cstring>

#include<iostream>

#include<algorithm>

#include<queue>

#include<stack>

#include<set>

using namespace std;

typedef long long ll;

typedef unsigned long long ull;

const int MAX_N=100000;

const int INF = 0x3f3f3f3f;

int n,m,k;

char map[1010][1010];

bool vis[1010][1010];



struct p{

	int x;

	int y;

	int step;

};

int ans=0;



bool yesorno(p x){

	if(x.x<1||x.x>m||x.y<1||x.y>n||map[x.y][x.x]=='X'||vis[x.y][x.x]==1)

		return 0;

	return 1;

}



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

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

void bfs(p st,int ww){

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

	if(ww==k)

		return;

	int temp=0;

	queue<p> q;

	q.push(st);

//	cout<<st.step<<" "<<ww<<" "<<k<<endl;

	while(!q.empty()){

		

		p ne=q.front();

		q.pop();

		if(map[ne.y][ne.x]==(char)'0'+ww+1){

			ans=ne.step;

			bfs(ne,ww+1);

			return;

		}

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

			p tem=ne;

			tem.x=tem.x+sx[i];

			tem.y=tem.y+sy[i];

			tem.step++;

			if(yesorno(tem)){

				q.push(tem);

				vis[tem.y][tem.x]=1;

			}

				

		}

	}

}

int main(){

	ios::sync_with_stdio(false);

	while(cin>>n>>m>>k){

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

		p st;

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

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

				cin>>map[i][j];

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

					st.x=j;

					st.y=i;

					st.step=0;

				}

			}

		}

		bfs(st,0);

		cout<<ans<<endl;

	}





	return 0;

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

#include<cstdio>

#include<cstring>

#include<iostream>

#include<algorithm>

#define SC(x) scanf("%d",&x);

#define pr(x)  cout << x << endl;

#define mem(arr,v) memset(arr,v,sizeof arr);

#define bug(x) cout << "#x = " << x <<endl;

const int inf = 0x3f3f3f3f;

const int maxn = 1010;

typedef long long ll;

using namespace std;

char Mz[maxn][maxn],c;

bool vis[maxn][maxn];

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

int M,N,C;

int px[10],py[10];

struct node {

	int x,y,step;

};



int bfs(int des,int sx,int sy) {

	mem(vis,0);

	vis[sx][sy] = 1;

	node nod = {sx,sy,0};

	queue<node> q;

	q.push(nod);



	while(!q.empty()) {

		nod =  q.front();

		q.pop();

		

		int x = nod.x, y = nod.y;

		int step =  nod.step;

		

		if(Mz[x][y] == des + '0')

			return step;

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

			int dx = x + dir[i][0];

			int dy = y + dir[i][1];

			if(Mz[dx][dy] == 'X' || Mz[dx][dy] == -1 || vis[dx][dy])	continue;

			nod = {dx,dy,step + 1};

			q.push(nod);

			vis[dx][dy] = 1;

		}

	}

}



int main() {

//	freopen("in.txt","r",stdin);

	cin >> N >> M >> C;

	mem(Mz,-1)

	mem(px,0)  mem(py,0)



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

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

			cin >> c;

			Mz[i][j] = c;

			if(c == '.' || c == 'X')	continue;

			if(c == 'S') px[0] = i, py[0] = j;

			else

				px[c - '0'] = i , py[c - '0'] = j;

		}

	int res = 0;

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

		if(px[i] && py[i]) {

//				dfs(i + 1,px[i],py[i],0);

			res += bfs(i + 1,px[i],py[i]);

		}

	}

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

}

// Faster C++ solution for the same problem (measured speedup: 1.3379176342713457x):
#include<cstdio>

#include<cstdlib>

#include<cmath>

#include<cstring>

#include<iostream>

#include<algorithm>

#include<queue>

#include<stack>

#include<set>

using namespace std;

typedef long long ll;

typedef unsigned long long ull;

const int MAX_N=100000;

const int INF = 0x3f3f3f3f;

int n,m,k;

char map[1010][1010];

bool vis[1010][1010];



struct p{

	int x;

	int y;

	int step;

};

int ans=0;



bool yesorno(p x){

	if(x.x<1||x.x>m||x.y<1||x.y>n||map[x.y][x.x]=='X'||vis[x.y][x.x]==1)

		return 0;

	return 1;

}



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

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

void bfs(p st,int ww){

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

	if(ww==k)

		return;

	int temp=0;

	queue<p> q;

	q.push(st);

//	cout<<st.step<<" "<<ww<<" "<<k<<endl;

	while(!q.empty()){

		

		p ne=q.front();

		q.pop();

		if(map[ne.y][ne.x]==(char)'0'+ww+1){

			ans=ne.step;

			bfs(ne,ww+1);

			return;

		}

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

			p tem=ne;

			tem.x=tem.x+sx[i];

			tem.y=tem.y+sy[i];

			tem.step++;

			if(yesorno(tem)){

				q.push(tem);

				vis[tem.y][tem.x]=1;

			}

				

		}

	}

}

int main(){

	ios::sync_with_stdio(false);

	while(cin>>n>>m>>k){

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

		p st;

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

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

				cin>>map[i][j];

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

					st.x=j;

					st.y=i;

					st.step=0;

				}

			}

		}

		bfs(st,0);

		cout<<ans<<endl;

	}





	return 0;

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

#include<cstring>

#include<queue>

using namespace std;

int map[1005][1005];

struct node{

	int x,y,time;

}s[10],p;

queue<node> q;

int vis[1005][1005];

int n,m,T;

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

int bfs(int a,int b){

	while(!q.empty())

		q.pop();

	q.push(s[a]);

	memset(vis,0,sizeof vis);

	vis[s[a].x][s[a].y]=1;

	while(!q.empty()){

		p=q.front();

		q.pop();

		int x=p.x,y=p.y,time=p.time;

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

			int x1=x+dir[i][0],y1=y+dir[i][1];

			if(x1<=0||y1<=0||x1>n||y1>m||vis[x1][y1]||map[x1][y1])

				continue ;

			if(x1==s[b].x&&y1==s[b].y)

				return time+1;

			vis[x1][y1]=1;

			p.x=x1,p.y=y1,p.time=time+1;

			q.push(p);

		}

	}

}

int main()

{

	scanf("%d%d%d",&n,&m,&T);

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

		scanf("\n");

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

			char c;

			c=getchar();

			if(c=='S')

				s[0].x=i,s[0].y=j;

			if(c=='X')

				map[i][j]=1;

			if(c>='1'&&c<='9')

				s[c-'0'].x=i,s[c-'0'].y=j;

		}

	}

	int time=0;

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

		time+=bfs(i,i+1);

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

}

// Faster C++ solution for the same problem (measured speedup: 2.2160632199294317x):
#include<cstdio>

#include<cstdlib>

#include<cmath>

#include<cstring>

#include<iostream>

#include<algorithm>

#include<queue>

#include<stack>

#include<set>

using namespace std;

typedef long long ll;

typedef unsigned long long ull;

const int MAX_N=100000;

const int INF = 0x3f3f3f3f;

int n,m,k;

char map[1010][1010];

bool vis[1010][1010];



struct p{

	int x;

	int y;

	int step;

};

int ans=0;



bool yesorno(p x){

	if(x.x<1||x.x>m||x.y<1||x.y>n||map[x.y][x.x]=='X'||vis[x.y][x.x]==1)

		return 0;

	return 1;

}



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

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

void bfs(p st,int ww){

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

	if(ww==k)

		return;

	int temp=0;

	queue<p> q;

	q.push(st);

//	cout<<st.step<<" "<<ww<<" "<<k<<endl;

	while(!q.empty()){

		

		p ne=q.front();

		q.pop();

		if(map[ne.y][ne.x]==(char)'0'+ww+1){

			ans=ne.step;

			bfs(ne,ww+1);

			return;

		}

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

			p tem=ne;

			tem.x=tem.x+sx[i];

			tem.y=tem.y+sy[i];

			tem.step++;

			if(yesorno(tem)){

				q.push(tem);

				vis[tem.y][tem.x]=1;

			}

				

		}

	}

}

int main(){

	ios::sync_with_stdio(false);

	while(cin>>n>>m>>k){

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

		p st;

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

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

				cin>>map[i][j];

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

					st.x=j;

					st.y=i;

					st.step=0;

				}

			}

		}

		bfs(st,0);

		cout<<ans<<endl;

	}





	return 0;

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

#include <cstdio>

#include <cstring>

#include <cmath>

#include <vector>

#include <queue>

#include <set>

#include <map>

#include <stack>

#include <utility>

#include <algorithm>

#include <sstream>

using namespace std;

typedef pair<int,int> P;



const int max_n=1000+10;

int h,w,n,sx,sy,dx[]={0,0,-1,1},dy[]={-1,1,0,0},d[max_n][max_n],ans;

char c[max_n][max_n];



void bfs(int x,int y,int m){

	if(m>n)return;

	queue<P> q;

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

	q.push(P(x,y));

	d[x][y]=1;

	while(!q.empty()){

		P p=q.front();

		q.pop();

		if(c[p.first][p.second]=='0'+m){

		    ans+=d[p.first][p.second]-1;

		    bfs(p.first,p.second,m+1);

		    return;

		}

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

			int tx=p.first+dx[i],ty=p.second+dy[i];

			if(tx>=0 && tx<h && ty>=0 && ty<w && c[tx][ty]!='X' && !d[tx][ty]){

				q.push(P(tx,ty));

				d[tx][ty]=d[p.first][p.second]+1;

			}

		}

	}

}



int main(){

    scanf("%d%d%d",&h,&w,&n);

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

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

            cin>>c[i][j];

            if(c[i][j]=='S'){sx=i;sy=j;}

        }

    bfs(sx,sy,1);

    cout<<ans<<endl;

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 2.273857415759427x):
#include<cstdio>

#include<cstdlib>

#include<cmath>

#include<cstring>

#include<iostream>

#include<algorithm>

#include<queue>

#include<stack>

#include<set>

using namespace std;

typedef long long ll;

typedef unsigned long long ull;

const int MAX_N=100000;

const int INF = 0x3f3f3f3f;

int n,m,k;

char map[1010][1010];

bool vis[1010][1010];



struct p{

	int x;

	int y;

	int step;

};

int ans=0;



bool yesorno(p x){

	if(x.x<1||x.x>m||x.y<1||x.y>n||map[x.y][x.x]=='X'||vis[x.y][x.x]==1)

		return 0;

	return 1;

}



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

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

void bfs(p st,int ww){

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

	if(ww==k)

		return;

	int temp=0;

	queue<p> q;

	q.push(st);

//	cout<<st.step<<" "<<ww<<" "<<k<<endl;

	while(!q.empty()){

		

		p ne=q.front();

		q.pop();

		if(map[ne.y][ne.x]==(char)'0'+ww+1){

			ans=ne.step;

			bfs(ne,ww+1);

			return;

		}

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

			p tem=ne;

			tem.x=tem.x+sx[i];

			tem.y=tem.y+sy[i];

			tem.step++;

			if(yesorno(tem)){

				q.push(tem);

				vis[tem.y][tem.x]=1;

			}

				

		}

	}

}

int main(){

	ios::sync_with_stdio(false);

	while(cin>>n>>m>>k){

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

		p st;

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

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

				cin>>map[i][j];

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

					st.x=j;

					st.y=i;

					st.step=0;

				}

			}

		}

		bfs(st,0);

		cout<<ans<<endl;

	}





	return 0;

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

#include <cstdio>

#include <cstring>

#include <cmath>

#include <vector>

#include <queue>

#include <set>

#include <map>

#include <stack>

#include <utility>

#include <algorithm>

#include <sstream>

using namespace std;



struct Node{

	int x,y;

};



const int max_n=1000+10;

int h,w,n,dx[]={0,0,-1,1},dy[]={-1,1,0,0},d[max_n][max_n],ans;

Node sN;

char c[max_n][max_n];



void bfs(int m){

	if(m>n)return;

	queue<Node> q;

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

	q.push(sN);

	d[sN.x][sN.y]=1;

	while(!q.empty()){

		Node p=q.front();

		q.pop();

		

		if(c[p.x][p.y]=='0'+m){

		    ans+=d[p.x][p.y]-1;

		    sN.x=p.x;

		    sN.y=p.y;

		    return;

		}

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

			Node nN;

			nN.x=p.x+dx[i];

			nN.y=p.y+dy[i];

			if(nN.x>=0 && nN.x<h 

            && nN.y>=0 && nN.y<w 

		   	&& c[nN.x][nN.y]!='X' && !d[nN.x][nN.y]){

				q.push(nN);

				d[nN.x][nN.y]=d[p.x][p.y]+1;

			}

		}

	}

}



int main(){

    scanf("%d%d%d",&h,&w,&n);

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

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

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

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

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

            	sN.x=i;

            	sN.y=j;

            	break;

            }

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

    cout<<ans<<endl;

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 2.2726940233979707x):
#include<cstdio>

#include<cstdlib>

#include<cmath>

#include<cstring>

#include<iostream>

#include<algorithm>

#include<queue>

#include<stack>

#include<set>

using namespace std;

typedef long long ll;

typedef unsigned long long ull;

const int MAX_N=100000;

const int INF = 0x3f3f3f3f;

int n,m,k;

char map[1010][1010];

bool vis[1010][1010];



struct p{

	int x;

	int y;

	int step;

};

int ans=0;



bool yesorno(p x){

	if(x.x<1||x.x>m||x.y<1||x.y>n||map[x.y][x.x]=='X'||vis[x.y][x.x]==1)

		return 0;

	return 1;

}



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

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

void bfs(p st,int ww){

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

	if(ww==k)

		return;

	int temp=0;

	queue<p> q;

	q.push(st);

//	cout<<st.step<<" "<<ww<<" "<<k<<endl;

	while(!q.empty()){

		

		p ne=q.front();

		q.pop();

		if(map[ne.y][ne.x]==(char)'0'+ww+1){

			ans=ne.step;

			bfs(ne,ww+1);

			return;

		}

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

			p tem=ne;

			tem.x=tem.x+sx[i];

			tem.y=tem.y+sy[i];

			tem.step++;

			if(yesorno(tem)){

				q.push(tem);

				vis[tem.y][tem.x]=1;

			}

				

		}

	}

}

int main(){

	ios::sync_with_stdio(false);

	while(cin>>n>>m>>k){

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

		p st;

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

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

				cin>>map[i][j];

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

					st.x=j;

					st.y=i;

					st.step=0;

				}

			}

		}

		bfs(st,0);

		cout<<ans<<endl;

	}





	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: 4.132582497016165x):
#include<cstdio>

#include<cstdlib>

#include<cmath>

#include<cstring>

#include<iostream>

#include<algorithm>

#include<queue>

#include<stack>

#include<set>

using namespace std;

typedef long long ll;

typedef unsigned long long ull;

const int MAX_N=100000;

const int INF = 0x3f3f3f3f;

int n,m,k;

char map[1010][1010];

bool vis[1010][1010];



struct p{

	int x;

	int y;

	int step;

};

int ans=0;



bool yesorno(p x){

	if(x.x<1||x.x>m||x.y<1||x.y>n||map[x.y][x.x]=='X'||vis[x.y][x.x]==1)

		return 0;

	return 1;

}



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

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

void bfs(p st,int ww){

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

	if(ww==k)

		return;

	int temp=0;

	queue<p> q;

	q.push(st);

//	cout<<st.step<<" "<<ww<<" "<<k<<endl;

	while(!q.empty()){

		

		p ne=q.front();

		q.pop();

		if(map[ne.y][ne.x]==(char)'0'+ww+1){

			ans=ne.step;

			bfs(ne,ww+1);

			return;

		}

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

			p tem=ne;

			tem.x=tem.x+sx[i];

			tem.y=tem.y+sy[i];

			tem.step++;

			if(yesorno(tem)){

				q.push(tem);

				vis[tem.y][tem.x]=1;

			}

				

		}

	}

}

int main(){

	ios::sync_with_stdio(false);

	while(cin>>n>>m>>k){

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

		p st;

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

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

				cin>>map[i][j];

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

					st.x=j;

					st.y=i;

					st.step=0;

				}

			}

		}

		bfs(st,0);

		cout<<ans<<endl;

	}





	return 0;

}
// Slower C++ solution:
#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;

}

// Faster C++ solution for the same problem (measured speedup: 2.2141631961959827x):
#include<cstdio>

#include<cstdlib>

#include<cmath>

#include<cstring>

#include<iostream>

#include<algorithm>

#include<queue>

#include<stack>

#include<set>

using namespace std;

typedef long long ll;

typedef unsigned long long ull;

const int MAX_N=100000;

const int INF = 0x3f3f3f3f;

int n,m,k;

char map[1010][1010];

bool vis[1010][1010];



struct p{

	int x;

	int y;

	int step;

};

int ans=0;



bool yesorno(p x){

	if(x.x<1||x.x>m||x.y<1||x.y>n||map[x.y][x.x]=='X'||vis[x.y][x.x]==1)

		return 0;

	return 1;

}



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

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

void bfs(p st,int ww){

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

	if(ww==k)

		return;

	int temp=0;

	queue<p> q;

	q.push(st);

//	cout<<st.step<<" "<<ww<<" "<<k<<endl;

	while(!q.empty()){

		

		p ne=q.front();

		q.pop();

		if(map[ne.y][ne.x]==(char)'0'+ww+1){

			ans=ne.step;

			bfs(ne,ww+1);

			return;

		}

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

			p tem=ne;

			tem.x=tem.x+sx[i];

			tem.y=tem.y+sy[i];

			tem.step++;

			if(yesorno(tem)){

				q.push(tem);

				vis[tem.y][tem.x]=1;

			}

				

		}

	}

}

int main(){

	ios::sync_with_stdio(false);

	while(cin>>n>>m>>k){

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

		p st;

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

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

				cin>>map[i][j];

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

					st.x=j;

					st.y=i;

					st.step=0;

				}

			}

		}

		bfs(st,0);

		cout<<ans<<endl;

	}





	return 0;

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

#include<cstring>

#include<queue>

using namespace std;

int n,m,k,step[1010][1010],temp[1010][3];

char a[1010][1010];

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

bool vis[1010][1010];

typedef pair<int,int> P;

int bfs(int x,int y,int ex,int ey)

{

	memset(vis,true,sizeof(vis));

	memset(step,0,sizeof(step));

	queue <P> que;

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

	vis[x][y]=false;

	while(!que.empty())

	{

		P tmp=que.front();que.pop();

		int b=que.size();

		if(tmp.first==ex&&tmp.second==ey)

		  return step[tmp.first][tmp.second];

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

		{

			int tx=tmp.first+dx[i];

			int ty=tmp.second+dy[i];

			if(tx>0&&tx<=n&&ty>0&&ty<=m&&vis[tx][ty]==true&&a[tx][ty]!='X')

			{

				que.push(P(tx,ty));

				vis[tx][ty]=false;

				step[tx][ty]=step[tmp.first][tmp.second]+1;

			}

		}

	}

}

int main()

{

	ios::sync_with_stdio(false);

	cin>>n>>m>>k;

		int sx,sy;

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

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

		  {

			cin>>a[i][j];

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

			  sx=i,sy=j;

			//temp[0][0]=i;

			//temp[0][1]=j;

		  }

		a[sx][sy]='0';

		int tot=0,ans=0;

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

		{

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

			  for(int l=1;l<=m;l++)

			    if(a[j][l]=='0'+i)

			    {

			    	tot=bfs(sx,sy,j,l);

			    	sx=j,sy=l;

				}

			ans+=tot;

		}

		cout<<ans<<endl;

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 2.7580653599832483x):
#include<cstdio>

#include<cstdlib>

#include<cmath>

#include<cstring>

#include<iostream>

#include<algorithm>

#include<queue>

#include<stack>

#include<set>

using namespace std;

typedef long long ll;

typedef unsigned long long ull;

const int MAX_N=100000;

const int INF = 0x3f3f3f3f;

int n,m,k;

char map[1010][1010];

bool vis[1010][1010];



struct p{

	int x;

	int y;

	int step;

};

int ans=0;



bool yesorno(p x){

	if(x.x<1||x.x>m||x.y<1||x.y>n||map[x.y][x.x]=='X'||vis[x.y][x.x]==1)

		return 0;

	return 1;

}



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

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

void bfs(p st,int ww){

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

	if(ww==k)

		return;

	int temp=0;

	queue<p> q;

	q.push(st);

//	cout<<st.step<<" "<<ww<<" "<<k<<endl;

	while(!q.empty()){

		

		p ne=q.front();

		q.pop();

		if(map[ne.y][ne.x]==(char)'0'+ww+1){

			ans=ne.step;

			bfs(ne,ww+1);

			return;

		}

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

			p tem=ne;

			tem.x=tem.x+sx[i];

			tem.y=tem.y+sy[i];

			tem.step++;

			if(yesorno(tem)){

				q.push(tem);

				vis[tem.y][tem.x]=1;

			}

				

		}

	}

}

int main(){

	ios::sync_with_stdio(false);

	while(cin>>n>>m>>k){

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

		p st;

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

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

				cin>>map[i][j];

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

					st.x=j;

					st.y=i;

					st.step=0;

				}

			}

		}

		bfs(st,0);

		cout<<ans<<endl;

	}





	return 0;

}