POJ：3259-Wormholes（最短路判断负环）

Wormholes

Time Limit: 2000MS Memory Limit: 65536K

Total Submissions: 58153 Accepted: 21747

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.

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解题心得：

1. 题意就是一个图里面有很多双向边，也有单向的负权路径，问有没有负权环。
2. 这里主要就是问了一个关于最短路中怎么判断是否存在负环，一般有负权路径的图都是用SPFA来跑，如果不存在负环一个点最多被压入队列中n次（被其余的每个点更新），那么如果大于n次说明存在负环，还是很容易想明白的，代码实现也很简单，记录一下压入队列的次数就行了。

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#include <stdio.h>
#include <cstring>
#include <vector>
#include <queue>
using namespace std;
const int maxn = 510;
int Time[maxn],n,m,w;
bool vis[maxn];
int maps[maxn][maxn];

void init() {
    memset(Time,0x3f, sizeof(Time));
    memset(maps,0x3f,sizeof(maps));
    scanf("%d%d%d",&n,&m,&w);
    for(int i=0;i<m;i++) {
        int a,b,c;
        scanf("%d%d%d",&a,&b,&c);
        if(maps[a][b] > c)
            maps[a][b] = maps[b][a] = c;
    }
    for(int i=0;i<w;i++) {
        int a,b,c;
        scanf("%d%d%d",&a,&b,&c);
        c = -c;
        if(maps[a][b] > c)
            maps[a][b] = c;
    }
}

bool spfa() {
    int num[maxn];
    memset(num,0, sizeof(num));
    memset(vis,0,sizeof(vis));
    num[1] = 1;
    queue <int> qu;
    qu.push(1);
    Time[1] = 0;
    while(!qu.empty()) {
        int now = qu.front(); qu.pop();
        vis[now] = false;
        for(int i=1;i<=n;i++) {
            if(maps[now][i] != 0x3f3f3f3f) {
                if(Time[now] + maps[now][i] < Time[i]) {
                    Time[i] = maps[now][i] + Time[now];
                    if(!vis[i]) {
                        vis[i] = true;
                        qu.push(i);
                        num[i]++;
                        if(num[i] > n) {
                            return true;
                        }
                    }
                }
            }
        }
    }
    return false;
}

int main() {
    int t;
    scanf("%d",&t);
    while(t--) {
        init();
        bool is_back = spfa();
        if(is_back)
            printf("YES\n");
        else
            printf("NO\n");
    }
    return 0;
}