2018 宁夏邀请赛 Moving On Gym - 102222F（深入理解三维floyd）

Firdaws and Fatinah are living in a country with n cities, numbered from 1 to n

. Each city has a risk of kidnapping or robbery.

Firdaws’s home locates in the city u
, and Fatinah’s home locates in the city v. Now you are asked to find the shortest path from the city u to the city v that does not pass through any other city with the risk of kidnapping or robbery higher than w

, a threshold given by Firdaws.

Input

The input contains several test cases, and the first line is a positive integer T

indicating the number of test cases which is up to 50

.

For each test case, the first line contains two integers n (1≤n≤200)
which is the number of cities, and q (1≤q≤2×104) which is the number of queries that will be given. The second line contains n integers r1,r2,⋯,rn indicating the risk of kidnapping or robbery in the city 1 to n respectively. Each of the following n lines contains n integers, the j-th one in the i-th line of which, denoted by di,j, is the distance from the city i to the city j

.

Each of the following q
lines gives an independent query with three integers u,v and w

, which are described as above.

We guarantee that 1≤ri≤105
, 1≤di,j≤105 (i≠j), di,i=0 and di,j=dj,i. Besides, each query satisfies 1≤u,v≤n and 1≤w≤105

.

Output

For each test case, output a line containing Case #x: at first, where x is the test case number starting from 1

. Each of the following q

lines contains an integer indicating the length of the shortest path of the corresponding query.

Example
Input

1
3 6
1 2 3
0 1 3
1 0 1
3 1 0
1 1 1
1 2 1
1 3 1
1 1 2
1 2 2
1 3 2

Output

Case #1:
0
1
3
0
1
2

#include
#include
#include
#include
using namespace std;
int dp[310][310][310];
int has[310];int dan[310];

int num[310];

bool cmp(int a,int b){
    return dan[a]<dan[b];
}


int main(){
	int t;
	int cur=1;
	scanf("%d",&t);
	while(t--){
		int n,m;
		scanf("%d%d",&n,&m);
		memset(dp,0x3f3f3f3f,sizeof(dp));
		int cnt=1;
		for(int i=1;i<=n;i++){
			scanf("%d",&dan[i]);
			num[i]=i;
		}
		for(int i=1;i<=n;i++)
			for(int j=1;j<=n;j++)
			scanf("%d",&dp[i][j][0]);
							
		sort(num+1,num+1+n,cmp);
		for(int k=1;k<=n;k++){
			for(int i=1;i<=n;i++){
				for(int j=1;j<=n;j++){
					dp[i][j][k]=min(dp[i][j][k-1],dp[i][num[k]][k-1]+dp[num[k]][j][k-1]);
				}
			}
		}
		printf("Case #%d:\n",cur++);
		while(m--){
			int st,ed,w;
			scanf("%d%d%d",&st,&ed,&w);
			int ans=0;
			for(int i=1;i<=n;i++){
				if(dan[num[i]]<=w) ans=i;
			}
			printf("%d\n",dp[st][ed][ans]);
			}
		}
}