任意两点之间的最短路（floyed)

F、Moving On

Firdaws and Fatinah are living in a country with nn cities, numbered from 11 to nn. Each city has a risk of kidnapping or robbery.

Firdaws's home locates in the city uu, and Fatinah's home locates in the city vv. Now you are asked to find the shortest path from the city uu to the city vv that does not pass through any other city with the risk of kidnapping or robbery higher than ww, a threshold given by Firdaws.

Input

The input contains several test cases, and the first line is a positive integer TT indicating the number of test cases which is up to 5050.

For each test case, the first line contains two integers n (1 \le n \le 200)n(1≤n≤200) which is the number of cities, and q (1 \le q \le 2 \times 10^4)q(1≤q≤2×104) which is the number of queries that will be given. The second line contains nn integers r_1, r_2, \cdots , r_nr1,r2,⋯,rn indicating the risk of kidnapping or robbery in the city 11 to nn respectively. Each of the following nn lines contains nn integers, the jj-th one in the ii-th line of which, denoted by d_{i,j}di,j, is the distance from the city iito the city jj.

Each of the following qq lines gives an independent query with three integers u,vu,v and ww, which are described as above.

We guarantee that 1 \le r_i \le 10^5, 1 \le d_{i,j} \le 10^5 (i \neq j), d_{i,i} = 01≤ri≤105,1≤di,j≤105(i=j),di,i=0 and d_{i,j} = d_{j,i}di,j=dj,i. Besides, each query satisfies 1 \le u,v \le n1≤u,v≤n and 1 \le w \le 10^51≤w≤105.

Output

For each test case, output a line containing Case #x: at first, where xx is the test case number starting from 11. Each of the following qq lines contains an integer indicating the length of the shortest path of the corresponding query.

输出时每行末尾的多余空格，不影响答案正确性

样例输入复制

样例输出复制

Case #1:

0

1

3

0

1

2

题意：输入t，表示t个测试样例
输入n,q,表示n个点，q次询问
下一行n个数表示[1,n]个点的危险值
接下来n行，每行3个数描述一条边，点x到y的距离为z;
最后q行，每行3个数，求从起点U到终点V,在每一个点危险值不超过W的前提下的最短距离

#include<iostream>

#include<string.h>

#include<string>

#include<algorithm>

#include<math.h>

#include<string>

#include<string.h>

#include<vector>

#include<utility>

#include<map>

#include<queue>

#include<set>

#define mx 0x3f3f3f3f

#define ll long long

#define MAXN 100

using namespace std;

int dp[][][];//dp[k][i][j]表示 添加(第1~k小的危险值的城市后)的 i->j 最短路

int vis[],rk[];

bool cmp(int i,int j)

{

    return rk[i]<rk[j];

}

int main()

{

    int t;

    scanf("%d",&t);

    for(int tt=;tt<=t;tt++)

    {

        memset(dp,mx,sizeof(dp));

        int n,q;

        scanf("%d%d",&n,&q);

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

        {

            vis[i]=i;//初始化排名

            scanf("%d",&rk[i]);

        }

        sort(vis+,vis+n+,cmp);//把城市的编号按风险等级从小到大排序，vis[排名]=编号

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

        //     cout<<vis[i]<<' ';

        // cout<<"<-----"<<endl;

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

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

                scanf("%d",&dp[][i][j]);//初始化每一条边的危险等级为0

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

        {

            int now=vis[k];

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

            {

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

                    dp[k][i][j]=min(dp[k-][i][j],dp[k-][i][now]+dp[k-][now][j]);

            }

        }

        printf("Case #%d:\n",tt);

        for(int i=;i<=q;i++)

        {

            int u,v,w;

            scanf("%d%d%d",&u,&v,&w);

            int k=;

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

                if(rk[vis[j]]<=w)//把危险值最接近w的边都加进去之后的最短路

                    k=j;

            printf("%d\n",dp[k][u][v]);

        }

    }

    return ;

}