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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 Format

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~(1le nle 200)n (1≤n≤200) which is the number of cities, and q~(1le qle 2 imes 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 nnlines 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 ii to 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 1le r_i le 10^51≤ri​≤105, 1le d_{i,j}le 10^5~(i eq j)1≤di,j​≤105 (i≠j), d_{i,i}=0di,i​=0 and d_{i,j}=d_{j,i}di,j​=dj,i​.Besides, each query satisfies 1le u,vle n1≤u,v≤n and 1le wle 10^51≤w≤105.

Output Format

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.

样例输入

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

样例输出

Case #1:
0
1
3
0
1
2

题目来源

The 2018 ACM-ICPC Chinese Collegiate Programming Contest

 

思路：动态规划f[i][j][k]记录经过前i个城市从j到k的最小风险，再将每个城市的风险进行排序，用于转移与计算。

代码：

#include"bits/stdc++.h"
#define ci(x) scanf("%d",&x)
const int N   = 1e6 + 5;
using namespace std;
int t,n,q;
struct P
{
    int id,r;
};
P a[N];
int f[205][205][205];
bool cmp(P a,P b){
    return a.r<b.r;
}
void cal(int m){
    for(int i=1;i<=m;i++){
        int id=a[i].id;
        for(int j=1;j<=n;j++){
            for(int k=1;k<=n;k++){
                f[i][j][k]=min(f[i-1][j][k],f[i-1][j][id]+f[i-1][id][k]);
            }
        }
    }
}
int main(){
    //freopen("data.in","r",stdin);
    //freopen("data.out","w",stdout);
    scanf("%d",&t);
    for(int ii=1;ii<=t;ii++){
        scanf("%d%d",&n,&q);
        for(int i=1;i<=n;i++) ci(a[i].r),a[i].id=i;
        memset(f,inf,sizeof(f));
        for(int i=1;i<=n;i++)
            for(int j=1;j<=n;j++) ci(f[0][i][j]);
        sort(a+1,a+n+1,cmp);
        cal(n);
        printf("Case #%d:
",ii);
        int u,v,w;
        for(int i=0;i<q;i++){
            ci(u),ci(v),ci(w);
            int k;
            for(k=1;k<=n;k++) if(a[k].r>w) break;
            printf("%d
",f[k-1][u][v]);
        }
    }
    return 0;
}