POJ #3259 Wormholes 判负环

Description

思路

#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;
}

原文地址：https://www.cnblogs.com/Bw98blogs/p/8450068.html