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
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
分析
先将各城市按危险值排序，在跑floyd算法的时候按照危险值排序后的顺序，dp[i][j][k]表示从城市i到城市j借助排序后的前k个城市可以取到的最小距离，所以状态转移方程就是dp[i][j][k] = min( dp[i][j][k-1] , dp[i][name[k]][k-1] + dp[name[k]][j][k-1])。
代码
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define inf 0x3f3f3f3f
#define maxn 300
int f[maxn][maxn][maxn];
int danger[maxn];
int name[maxn];bool cmp(int a,int b){
return danger[a]<danger[b];
}int main(){
int i, j, k, n, m, T, q;int cur = 0;scanf("%d",&T);while(T--){
scanf("%d %d",&n,&q);for(i = 1; i <= n; i++){
name[i] = i;scanf("%d",&danger[i]);}sort(name+1, name+1+n, cmp);for(i = 1; i <= n; i++){
for(j = 1; j <= n; j++){
scanf("%d",&f[i][j][0]);}}for(k = 1; k <= n; k++){
for(i = 1; i <= n; i++){
for(j = 1; j <= n; j++){
f[i][j][k] = min(f[i][j][k-1], f[i][name[k]][k-1]+f[name[k]][j][k-1]);}}}printf("Case #%d:\n",++cur);while(q--){
int u, v ,w;scanf("%d%d%d",&u,&v,&w);int ans = 0;for(i = 1; i <= n; i++){
if(danger[name[i]]<=w) ans = i;}printf("%d\n",f[u][v][ans]);}}return 0;
}