Poj(3259)，SPFA，判负环

题目链接：http://poj.org/problem?id=3259

Wormholes

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 44090		Accepted: 16203

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.

 

题意：2个案例，3个点，3条路，1个虫洞，路是双向的，虫洞是单向的，单向的路为负，起点为1，看能不能穿越。

分析：SPFA判负环。

没有看模板，直接上代码，结果发现不知道WA了多少次，实在搞不下去了，最后，佳鑫大神帮我Debug了，我在建图的时候head初始化了，结果在SPFA里面有初始化为-1了，真是害死我了，最后还是RE了一次，数组开小了。不管怎么样，Dijkstra，SPFA，Floyd，Kruskal都可以不用看模板了，很开心。

#include <stdio.h>
#include<iostream>
#include <queue>
#include <string.h>

using namespace std;

#define INF 0x3f3f3f3f

int n,m,t;

struct Edge
{
    int v;
    int w;
    int next;
} edge[];

int NE = ;

int dis[];
int head[];
bool vis[];
int cnt[];

void add(int u,int v,int d)
{
    edge[NE].v = v;
    edge[NE].w = d;
    edge[NE].next = head[u];
    head[u]= NE++;
}

bool SPFA()
{
    for(int i=; i<=n; i++)
    {
        dis[i] = INF;
        vis[i] = false;
        cnt[i] = ;
    }

    dis[] = ;
    vis[] = true;
    cnt[] = ;
    queue<int>  Q;
    Q.push();

    while(!Q.empty())
    {
        int u=Q.front();
        Q.pop();
        vis[u]=false;
        for(int i=head[u]; i!=-; i=edge[i].next)
        {

            if(dis[edge[i].v]>dis[u]+edge[i].w)
            {
                dis[edge[i].v] = dis[u] + edge[i].w;
                if(!vis[edge[i].v])
                {
                    vis[edge[i].v] = true;
                    Q.push(edge[i].v);
                    if(++cnt[edge[i].v]>n)
                        return true;
                }
            }
        }
    }
    return false;
}

int main()
{
    //freopen("input.txt","r",stdin);
    int cases;
    scanf("%d",&cases);
    while(cases--)
    {

        NE = ;
        memset(head,-,sizeof(head));
        scanf("%d%d%d",&n,&m,&t);

        for(int i=; i<m; i++)
        {
            int a,b,d;
            scanf("%d%d%d",&a,&b,&d);
            add(a,b,d);
            add(b,a,d);
        }

        for(int i=; i<t; i++)
        {
            int a,b,d;
            scanf("%d%d%d",&a,&b,&d);
            add(a,b,-d);
        }

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