Problem:
Score : 400 points

Problem Statement

There are N towns in the State of Atcoder, connected by M bidirectional roads.

The i-th road connects Town A_i and B_i and has a length of C_i.

Joisino is visiting R towns in the state, r_1,r_2,..,r_R (not necessarily in this order).

She will fly to the first town she visits, and fly back from the last town she visits, but for the rest of the trip she will have to travel by road.

If she visits the towns in the order that minimizes the distance traveled by road, what will that distance be?

Constraints

2≤N≤200

1≤M≤N×(N-1)/2

2≤R≤min(8,N) (min(8,N) is the smaller of 8 and N.)

r_i≠r_j (i≠j)

1≤A_i,B_i≤N, A_i≠B_i

(A_i,B_i)≠(A_j,B_j),(A_i,B_i)≠(B_j,A_j) (i≠j)

1≤C_i≤100000

Every town can be reached from every town by road.

All input values are integers.

Input

Input is given from Standard Input in the following format:

N M R
r_1 ... r_R
A_1 B_1 C_1
:
A_M B_M C_M

Output

Print the distance traveled by road if Joisino visits the towns in the order that minimizes it.

Sample Input 1

3 3 3
1 2 3
1 2 1
2 3 1
3 1 4

Sample Output 1

2

For example, if she visits the towns in the order of 1, 2, 3, the distance traveled will be 2, which is the minimum possible.

Sample Input 2

3 3 2
1 3
2 3 2
1 3 6
1 2 2

Sample Output 2

4

The shortest distance between Towns 1 and 3 is 4. Thus, whether she visits Town 1 or 3 first, the distance traveled will be 4.

Sample Input 3

4 6 3
2 3 4
1 2 4
2 3 3
4 3 1
1 4 1
4 2 2
3 1 6

Sample Output 3

3

Solution in C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>
//#define ERROR -1

void fl(int M,int dp[M][M]);

int next_perm(int *p, int n)
{
  int i, j, k, tmp;

  /* 上位桁のほうが下位桁よりも小さいところまで移動 */
  for(i = n - 1; i > 0 && p[i-1] >= p[i]; i--);

  /* pが最大(次の並べ替えがない) */
  if(i == 0) return 0;

  /* p[i-1]より値の大きい最も下位の桁をp[j]とする */
  for(j = n - 1; j > i && p[i-1] >= p[j]; j--);

  /* p[i-1]とp[j]とを交換 */
  tmp = p[i-1], p[i-1] = p[j], p[j] = tmp;

  /* p[i]から最下位までを逆順 */
  for(k = 0; k <= ((n-1)-i)/2; k++)
    tmp = p[i+k], p[i+k] = p[(n-1)-k], p[(n-1)-k] = tmp;

  return 1;
}

int main(){
	int N,M,R;//N=街の数 M=道の数 R=訪れる街の数
	int i,j;//ループ用変数
	scanf("%d %d %d",&N,&M,&R);
/*	if(N<2||N>200)return ERROR;
	if(M<1||M>N*(N-1)/2)return ERROR;
	if(R<2||R>8||R>N)return ERROR;*/
	int res=0,res_tmp[50000],R_ka=1;
	int R_num[R];
	int dist[M];//キョリ
	int d[N][N];//どこからどこへ？のキョリ格納用
	int a,b,tmp;//一時変数

	for(i=0;i<N+1;i++){
		for(j=1;j<N+1;j++){
			d[i][j]=65535;
		}
	}

	//if(N<R)return ERROR;
	for(i=0;i<R;i++){
		scanf("%d ",&R_num[i]);
	}
	for(i=0;i<M;i++){
		scanf("%d %d %d",&a,&b,&dist[i]);
		//if(a==b||a==0||b==0||dist==0)return ERROR;
		d[a][b]=dist[i];//ex:a=1,b=2,dist=2→d[1][2]=2
	}
	for(i=1;i<N+1;i++){
		for(j=1;j<N+1;j++){
			if(d[i][j]!=65535&&d[j][i]==65535){
				d[j][i]=d[i][j];
			}
			else if(d[i][j]!=65535&&d[j][i]!=65535){
				if(d[i][j]>d[j][i]){
					d[i][j]=d[j][i];
				}else{
					d[j][i]=d[i][j];
				}
			}
		}
	}

	/*for(i=1;i<=N;i++){
		for(j=1;j<=N;j++){
			printf("%dと%dの最短距離＝%d\n",i,j,d[i][j]);
		}
	}*/
	fl(N,d);
    for(i=0;i<N+1;i++){
    	for(j=0;j<N+1;j++){
			if(d[i][j]==65535){
				d[i][j]=-1;
			}
			if(i==j){
				d[i][i]=-1;
			}
    	}
    }

    for(i=R;i!=1;i--){
    	R_ka=i*R_ka;
    }

    //printf("R!=%d\n",R_ka);
    for(j=0;j<R_ka;){
    	  /* 並べ替え */
    	  do {
    		  res=0;
    		  for(i=0;i<R-1;i++){
    			  res+=d[R_num[i]][R_num[i+1]];
    		  }
   			res_tmp[j]=res;
   			j++;
    	  } while(next_perm(R_num, R));
		/*for(i=0;i<R-1;i++){
			res+=d[R_num[i]][R_num[i+1]];
		}*/
		/*tmp=R_num[0];
		for(i=0;i<R;i++){
			if(i==R-1){
				R_num[i]=tmp;
			}else{
				R_num[i]=R_num[i+1];
			}
		}*/
    }
	/*for(i=1;i<=N;i++){
		for(j=1;j<=N;j++){
			printf("%dと%dの最短距離＝%d\n",i,j,d[i][j]);
		}
	}*/
    for(i=0;i<R_ka+1;i++){
    	if(res>res_tmp[i]){
    		res=res_tmp[i];
    	}
    }
    /*for(i=0;i<R_ka;i++){
    	printf("res_tmp=%d\n",res_tmp[i]);
    }*/
    printf("%d\n",res);
    return 0;
}

void fl(int M,int dp[M][M]){
	int i,j,k;
    for (k = 1; k < M+1; k++) {
      for (j = 1; j < M+1; j++) {
    	  for(i=1;i<M+1;i++){
			if (dp[i][j] > dp[i][k] + dp[k][j]){
			  dp[i][j] = dp[i][k] + dp[k][j];
			}
        }
      }
    }
}