POJ #3259 Wormholes 判负环 SPFA

Description

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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.

 

思路

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　　裸题，判断图是否有负环，用 bellman_ford 或者 spfa 。

#include<iostream>
#include<algorithm>
#include<vector>
#include<queue>
#include<cstring>
using namespace std;
#define INF 0x3f3f3f3f
int N, M, W; //顶点数 正环数 负权边数
const int MAX_N = 505;

struct Edge {
    int to;
    int T;
    Edge(int tt, int TT) : to(tt), T(TT) {}
};
vector<Edge> G[MAX_N];

void addEdge (const int& u, const int& v, const int& T) {
    G[u].push_back(Edge(v, T));
}

int dis[MAX_N];
bool inQueue[MAX_N];
int cnt[MAX_N];
bool spfa (const int& s) {
    memset(dis, INF, sizeof(dis));
    memset(inQueue, false, sizeof(inQueue));
    memset(cnt, 0, sizeof(cnt));
    dis[s] = 0;
    queue<int> q;
    q.push(s);
    inQueue[s] = true;
    while(!q.empty()) {
        int u = q.front(); q.pop();
        inQueue[u] = false;
        for (int j = 0; j < G[u].size(); ++j) {
            int v = G[u][j].to;
            int cost = G[u][j].T;
            if (dis[v] > dis[u] + cost ) {
                dis[v] = dis[u] + cost;
                if (!inQueue[v]) {
                    q.push(v);
                    inQueue[v] = true;
                    if (++cnt[v] > N) return false;
                }
            }
        }
    }
    return true;
}

int main(void) {
    int F;
    cin >> F;
    while (F--) {
        cin >> N >> M >> W;
        for (int i = 1; i <= N; i++) G[i].clear();
        for (int i = 1; i <= M; i++) {
            int S, E, T;
            cin >> S >> E >> T;
            addEdge (S, E, T);
            addEdge (E, S, T);
        }
        for (int i = 1; i <= W; i++) {
            int S, E, T;
            cin >> S >> E >> T;
            addEdge (S, E, -1 * T);
        }
        if (!spfa(1)) cout << "YES" << endl;
        else cout << "NO" << endl;
    }
    return 0;
}