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POJ3259 Wormholes 【SPFA判断负环】

2024-08-06 18:44:13   219人阅读

Wormholes
Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 32111 Accepted: 11662

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.

Source

USACO 2006 December Gold

判断图中是否存在负环。

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <queue>
using namespace std;
typedef long long LL;

#define maxn 505
#define maxm 5210
#define inf 0x3f3f3f3f

int T, N, W, M;
int head[maxn], id;
struct Node {
    int v, w, next;
} E[maxm];
int out[maxn], dist[maxn];
bool vis[maxn];

void addEdge(int u, int v, int w) {
    E[id].v = v; E[id].w = w;
    E[id].next = head[u]; 
    head[u] = id++;
}

void getMap() {
    memset(head, -1, sizeof(int) * (N + 1));
    int u, v, w; id = 0;
    while(M--) {
        scanf("%d%d%d", &u, &v, &w);
        addEdge(u, v, w);
        addEdge(v, u, w);
    }
    while(W--) {
        scanf("%d%d%d", &u, &v, &w);
        addEdge(u, v, -w);
    }
}

bool SPFA() {
    int i, j, u, v, w;
    queue<int> Q;
    for(i = 1; i <= N; ++i) {
        vis[i] = 1; out[i] = 0;
        dist[i] = 0; Q.push(i);
    }
    while(!Q.empty()) {
        u = Q.front(); Q.pop(); 
        if(++out[u] > N)
            return false; 
        vis[u] = 0;
        for(i = head[u]; i != -1; i = E[i].next) {
            v = E[i].v; w = E[i].w;
            if(dist[v] > dist[u] + w) {
                dist[v] = dist[u] + w;
                if(!vis[v]) {
                    vis[v] = 1;
                    Q.push(v);
                }
            }
        }
    }
    return true;
}

int main() {
    // freopen("stdin.txt", "r", stdin);
    scanf("%d", &T);
    while(T--) {
        scanf("%d%d%d", &N, &M, &W);
        getMap();
        printf(SPFA() ? "NO\n" : "YES\n");
    }
    return 0;
}


POJ3259 Wormholes 【SPFA判断负环】


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