（赛后补题）Gym - 101020 H H

今天模拟赛我给忘了，睡过了......来的时候已经过了一个小时了，然后就没做到这个题，其实特别简单。

ou and your friends are going to camping next weekend , because you are the best driver and you are living in this town for ten years , you have to drive the car from your home and pick up each of your friends to take them with you ,then finally go to the camping site. You have the map of the town and the length of every street ,and homes of friends will be at some intersections. you want to start from your home then pick up all of your friends then go to the camping site as fast as you can so you have to find the shortest path to do that. you can pick up your friends in any order but you have to make sure you pick up all of them. NOTE: you can pass by any intersection more than once including home and camping site .

Input

Your program will be tested on one or more test cases. The first line of the input will be a single integer T, the number of test cases (1 ≤ T ≤ 50). Followed by the test cases, the first line of each test case contains three integers N,M,F (3 ≤ N ≤ 100 , 0<M ≤ (N*(N-1))/2 , 1 ≤ F ≤ 9 ) representing the number of intersections , the number of streets and number of friends respectively. Followed by M lines, each line contains 3 integers separated by a single space X Y Z (1 ≤ X, Y ≤ N), (1 ≤ Z ≤ 1000) which represent a street between intersection X and intersection Y with length Z in both direction (X and Y will be different). then follow F (1<fr<N) deferent integer representing friends home number(number of intersection where home is located). your home is at intersection number one and camping site is at intersection number N.

Output

For each test case print a single line containing "Case n:" (without quotes) where n is the test case number (starting from 1) followed by a space then the shortest path to go from your home to camping site after picking up all your friends .

Sample 1
Inputcopy Outputcopy
2
4 4 2
1 4 3
1 2 2
2 3 2
3 4 2
2 3
6 8 2
1 2 1
2 3 2
3 6 3
1 4 11
4 5 2
5 6 20
5 3 5
4 3 10
4 5
Case 1: 6
Case 2: 20

Inputcopy	Outputcopy
2 4 4 2 1 4 3 1 2 2 2 3 2 3 4 2 2 3 6 8 2 1 2 1 2 3 2 3 6 3 1 4 11 4 5 2 5 6 20 5 3 5 4 3 10 4 5	Case 1: 6 Case 2: 20

题意：你是一个司机，你家在1路口，你要接你的几个朋友到路口露营，问最短路

思路：首先，看到n的范围比较小，然后m比较大，所以是稠密图，首先想到Floyd，因为Floyd是n^3的时间复杂度，在这里n是100，所以不会超时，而且它可以直接记录点i到点j的最短路径，所以我们可以用dfs遍历一下所有接送朋友的顺序的情况，然后有用minn记录最短距离。

话不多说，看代码：

#include <bits/stdc++.h>
using namespace std;
#define Maxn 0x3f3f3f3f
int mp[110][110];
int a[110];
int vis[110];
int n,m,num,minn;
void floy()//Floyd代码
{int i,j,k;for(k=1;k<=n;k++){for(i=1;i<=n;i++){for(j=1;j<=n;j++){mp[i][j]=min(mp[i][j],mp[i][k]+mp[k][j]);}}}
}
void dfs(int dis,int last,int level)//level记录这是第几个朋友，dis表示当前司机走过的距离//last表示上一个朋友所在的路口
{if(level>num){minn=min(minn,dis+mp[last][n]);return;}for(int i=1;i<=num;i++){if(vis[i]==0){vis[i]=1;dis+=mp[last][a[i]];dfs(dis,a[i],level+1);dis-=mp[last][a[i]];vis[i]=0;}}
}
void chushi()//Floyd初始化
{for(int i=1;i<=n;i++){for(int j=1;j<=n;j++){if(i==j)mp[i][j]=0;elsemp[i][j]=Maxn;}}
}int main()
{int t,x,y,z;scanf("%d",&t);for(int k=1;k<=t;k++){memset(vis,0,sizeof(vis));scanf("%d %d %d",&n,&m,&num);chushi();minn=Maxn;while(m--){scanf("%d %d %d",&x,&y,&z);mp[x][y]=min(z,mp[x][y]);mp[y][x]=min(z,mp[y][x]);}for(int i=1;i<=num;i++){scanf("%d",&a[i]);}floy();dfs(0,1,1);printf("Case %d: %d\n",k,minn);}return 0;
}

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