Poj 3259 Wormholes【SPFA判断负权回路】

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探讨了在一个包含特殊路径和虫洞的农场网络中是否存在能够使农民John返回到出发前时间的循环路径。通过Bellman-Ford和SPFA算法判断是否存在负权回路，从而实现时间旅行。

Wormholes

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 42123		Accepted: 15483

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

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

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.

Source

USACO 2006 December Gold

题目大意：

有t组数据，没组数据第一行三个数n，m，w，表示有n个节点，m条无向边，w条有向负权边，问是否有负权回路存在。

思路：

Bellman-Ford/SPFA算法都能用来判断负权回路是否存在。这里我的代码是SPFA代码实现的，设定一个出队次数数组out【i】，表示节点i的出队次数，如果某个节点的出队次数达到了n次，那么就说明这个图一定有负权回路存在。

AC代码：

#include<stdio.h>
#include<string.h>
#include<queue>
using namespace std;
int head[1000000];
struct EdgeNode
{
    int to;
    int w;
    int next;
}e[1000000];
int out[1000000];
int vis[1000000];
int dis[1000000];
int n,m,w,cont;
void add(int from,int to,int w)
{
    e[cont].to=to;
    e[cont].w=w;
    e[cont].next=head[from];
    head[from]=cont++;
}
int SPFA(int ss)
{
    memset(out,0,sizeof(out));
    memset(vis,0,sizeof(vis));
    vis[ss]=1;
    for(int i=1;i<=n;i++)dis[i]=0x3f3f3f3f;
    dis[ss]=0;
    queue<int >s;
    s.push(ss);
    while(!s.empty())
    {
        int u=s.front();
        s.pop();vis[u]=0;
        out[u]++;
        if(out[u]>=n)return 0;
        for(int j=head[u];j!=-1;j=e[j].next)
        {
            int v=e[j].to;
            int w=e[j].w;
            if(dis[v]>dis[u]+w)
            {
                dis[v]=dis[u]+w;
                if(vis[v]==0)
                {
                    vis[v]=1;
                    s.push(v);
                }
            }
        }
    }
    return 1;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        cont=0;
        memset(head,-1,sizeof(head));
        scanf("%d%d%d",&n,&m,&w);
        while(m--)
        {
            int x,y,w;
            scanf("%d%d%d",&x,&y,&w);
            add(x,y,w);
            add(y,x,w);
        }
        while(w--)
        {
            int x,y,w;
            scanf("%d%d%d",&x,&y,&w);
            add(x,y,-w);
        }
        int tt=SPFA(1);
        if(tt==1)
        {
            printf("NO\n");
        }
        else printf("YES\n");
    }
}