spfa判断负环

#include <bits/stdc++.h>
using namespace std;
const int N = 10010;

int h[N], e[N], ne[N], w[N], idx;
int dist[N], cnt[N];
bool st[N];
int n, m;

void add(int a, int b, int c) {
    e[idx] = b, w[idx] = c, ne[idx] = h[a], h[a] = idx++;
}

bool spfa() {
    queue<int> q;
    for(int i = 1; i <= n; i++) {
        q.push(i);
        st[i] = true;
    }
    while(q.size()) {
        auto t = q.front();
        q.pop();
        st[t] = false;
        for(int i = h[t]; i != -1; i = ne[i]) {
            int j = e[i];
            if(dist[j] > dist[t] + w[i]) {
                dist[j] = dist[t] + w[i];
                cnt[j] = cnt[t] + 1; // 如果从1号点到x的最短路中包含至少n个点(不包括自己),则说明存在环
                if(cnt[j] >= n) return true;
                if(!st[j]) {
                    q.push(j);
                    st[j] = true;
                }
            }
        }
    }
    return false;
}

int main() {
    scanf("%d%d",&n,&m);
    memset(h,-1,sizeof h);
    while(m -- ) {
        int a, b, c;
        scanf("%d%d%d",&a,&b,&c);
        add(a,b,c);
    }
    bool t = spfa();
    if(t) puts("Yes");
    else puts("No");
    return 0;
}
原文地址:https://www.cnblogs.com/yonezu/p/13492708.html