3259 spfa判断负环（邻接表）

Wormholes

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 44174		Accepted: 16235

Description

While exploring his many farms, Farmer John has discovered a number of amazing wormholes. A wormhole is very peculiar because it is a one-way path that delivers you to its destination at a time that is BEFORE you entered the wormhole! Each of FJ's farms comprises N (1 ≤ N ≤ 500) fields conveniently numbered 1..N, M (1 ≤ M ≤ 2500) paths, and W (1 ≤ W ≤ 200) wormholes.

As FJ is an avid time-traveling fan, he wants to do the following: start at some field, travel through some paths and wormholes, and return to the starting field a time before his initial departure. Perhaps he will be able to meet himself :) .

To help FJ find out whether this is possible or not, he will supply you with complete maps to F (1 ≤ F ≤ 5) of his farms. No paths will take longer than 10,000 seconds to travel and no wormhole can bring FJ back in time by more than 10,000 seconds.

Input

Line 1: A single integer, F. F farm descriptions follow.
Line 1 of each farm: Three space-separated integers respectively: N, M, and W
Lines 2..M+1 of each farm: Three space-separated numbers (S, E, T) that describe, respectively: a bidirectional path between S and E that requires T seconds to traverse. Two fields might be connected by more than one path.
Lines M+2..M+W+1 of each farm: Three space-separated numbers (S, E, T) that describe, respectively: A one way path from S to E that also moves the traveler back T seconds.

Output

Lines 1..F: For each farm, output "YES" if FJ can achieve his goal, otherwise output "NO" (do not include the quotes).

Sample Input

2
3 3 1
1 2 2
1 3 4
2 3 1
3 1 3
3 2 1
1 2 3
2 3 4
3 1 8

Sample Output

NO
YES

Hint

For farm 1, FJ cannot travel back in time.
For farm 2, FJ could travel back in time by the cycle 1->2->3->1, arriving back at his starting location 1 second before he leaves. He could start from anywhere on the cycle to accomplish this.

题目大意：

一个农民有农场，上面有一些虫洞和路，走虫洞可以回到 T秒前，而路就和平常的一样啦，需要花费时间走过。问该农民可不可能从某个点出发后回到该点，并且于出发时间之前?

直接上spfa，看了很多模板，只有这个模板的命名规范些，以后就照这个模板来做吧。

题目大意：
一个农民有农场，上面有一些虫洞和路，走虫洞可以回到 T秒前，而路就和平常的一样啦，需要花费时间走过。问该农民可不可能从某个点出发后回到该点，并且于出发时间之前?

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;
const int inf = 0x3f3f3f3f;
const int maxn = 520 ;
const int maxm = 5200;

struct edge
{
    int to;
    int val;
    int next;
}e[maxm];

int len,head[maxn];
int dis[maxn];
int n,m,c;

bool spfa()
{
    for(int i = 1;i<=n;i++)
        dis[i]=inf;

        bool vis[maxn] = {0};
        int cnt[maxn] = {0};
      //因为这里是找环，不是找到某点的最短路径，所以不用初始dis[start]=0;直接从1开始找。
        int cur = 1; //当前点初始化
        queue<int> q;
        q.push(cur);
        vis[cur] = true;
        cnt[cur]=1;
        dis[cur]=0; //以当前点的为始点的最短路径

        while(!q.empty())
        {
            cur=q.front();
            q.pop();
            vis[cur]=false;//出队则取消标记

            for(int i = head[cur];i!=-1;i=e[i].next)
            {
                int tto = e[i].to;
                if(  dis[tto]>dis[cur]+e[i].val )
                {
                    dis[tto] = dis[cur]+e[i].val;
                    if(!vis[tto]){
                        cnt[tto]++;
                        vis[tto]=true;
                        q.push(tto);//新点进队
                        if(cnt[cur]>n)
                            return true;//如果进队超过n次，则存在负环
                    }
            }
        }
    }
        return false;
}

void add(int from,int to,int val){
     e[len].to=to;
     e[len].val=val;
     e[len].next=head[from];
     head[from]=len++;
     }

int main(){
    int T;
    scanf("%d",&T);
    while(T--){
        memset(head,-1,sizeof(head));
        len=0;

        scanf("%d%d%d",&n,&m,&c);

        for(int i = 0;i<m;i++){
            int from,to,val;
        scanf("%d%d%d",&from,&to,&val);
        add(from,to,val);
        add(to,from,val);
    }

    for(int i = 0;i<c;i++)
    {
        int from,to,val;
        scanf("%d%d%d",&from,&to,&val);
        add(from,to,-val);
    }

    if(spfa())
        printf("YES");
    else
        printf("NO");
        printf("\n");
    }
    return 0;
    }