【Hautoj 1278 Transmit information】+ dfs

1278: Transmit information
时间限制: 3 秒 内存限制: 128 MB
提交: 24 解决: 13
提交 状态
题目描述
The Chinese people threw themselves into an all-out war of resistance against Japanese aggression in 1937. The first line of resistance against aggression was formed by spies and underground workers.
They lurked in every place of the city.

There’s a piece of information that needs to be passed on to them. Now there is a traffic map, each road connects two different intersections Xi and Yi, each of which is the termination for at least two road. The length of each road is known LENi, no two intersections are directly connected by two different roads.

N spies lurk at every intersection , some intersections mignt have more than one spy. For security, they must position themselves properly , each spy cannot accept information from local intersection , can only be transferred from elsewhere and end up at the finishing pace.

At first, the information is in the hands of a spy at S intersection. After N spies transmission, and finally arrived at E intersection .

Write a program to find the shortest path that connects the starting intersection(S) and the ending intersection(E) ang transmission exactly N spies.
输入
The first line of the input contains one integers T, which is the nember of test cases (1<=T<=5). Each test case specifies:

• Line 1: Four space-separated integers: N M S E

• Lines 2..M+1: Line i+1 describes road i with three space-separated integers: LENi Xi Yi

    (  2<=N<=300,000  2<=M<=100,  1<= LENi, Xi, Yi ,S, E <=1000  i=1,…,m) 
  

  输出
  For each test case generate a single line containing a single integer that is the shortest path from intersection S to intersection E that transmits exactly N spies.
  样例输入
  1
  2 6 6 4
  11 4 6
  4 4 8
  8 4 9
  6 6 8
  2 6 9
  3 8 9
  样例输出
  10
  提示

暴力 dfs

AC代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
struct node{
    int to,vl,next;
}st[210];
int head[1010],vis[1010],n,num,ans;
void add(int a,int b,int vl){
   st[num].vl = vl,st[num].to = b,st[num].next = head[a],head[a] = num++;
}
void dfs(int a,int cut,int en,int sum){
    for(int i = head[a]; i ;i = st[i].next){
        int o = st[i].to;
        if(!vis[o]){
            vis[o] = 1;
            if(cut + 1 == n && en == o) ans = min(ans,sum + st[i].vl);
            else if(cut + 1 < n) dfs(o,cut + 1,en,sum + st[i].vl);
            vis[o] = 0;
        }
    }
}
int main()
{
    int T,m,s,e;
    scanf("%d",&T);
    while(T--){
        num = 0;
        scanf("%d %d %d %d",&n,&m,&s,&e);
        while(m--){
           int vl,a,b; scanf("%d %d %d",&vl,&a,&b);
           add(a,b,vl),add(b,a,vl);
        }
        ans = 0x3f3f3f;
        vis[s] = 1,dfs(s,0,e,0);
        printf("%d\n",ans);
        memset(vis,0,sizeof(vis));
        memset(head,0,sizeof(head));
    }
    return 0;
}