AcWing 904. 虫洞 (spfa求负环模板)

题目链接：904. 虫洞 - AcWing题库

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题目大意：

约翰有 F 个农场，每个农场有 N 片田地、M 条双向路径和 W 个单向虫洞。
虫洞可以让你回到过去。
判断农场中是否存在负权回路，有输出 YES，否输出 NO。

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SPFA代码：

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<vector>
#include<queue>
#include<deque>
#include<stack>
#include<set>
#include<map>
#include<unordered_set>
#include<unordered_map>
#include<bitset>
#include<tuple>
#define inf 72340172838076673
#define int long long
#define endl '\n'
#define F first
#define S second
#define  mst(a,x) memset(a,x,sizeof (a))
using namespace std;
typedef pair<int, int> pii;

const int N = 200086, mod = 998244353;

int n, M, W;
int h[N], e[N], ne[N], w[N], idx;

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

bool spfa() {

    vector<int> dis(N);
    vector<int> cnt(N);
    vector<bool> st(N);
    stack<int> q;
    
    for (int i = 1; i <= n; i++) {
        q.push(i);
        st[i] = true;
    }
    
    while (q.size()) {
        int u = q.top();
        q.pop();
        st[u] = false;
        
        for (int i = h[u]; ~i; i = ne[i]) {
            int j = e[i];
            
            if (dis[j] > dis[u] + w[i]) {
                dis[j] = dis[u] + w[i];
                cnt[j] = cnt[u] + 1;
                if (cnt[j] >= n) return true;
                
                if (!st[j]) {
                    q.push(j);
                    st[j] = true;
                }
            }
        }
    }
    
    return false;
}


void solve() {
    mst(h, -1), idx = 0;
    cin >> n >> M >> W;
    for (int i = 1; i <= M; i++) {
        int a, b, c;
        cin >> a >> b >> c;
        add(a, b, c);
        add(b, a, c);
    }
    
    for (int i = 1; i <= W; i++) {
        int a, b, c;
        cin >> a >> b >> c;
        add(a, b, -c);
    }
    
    if (spfa()) cout << "YES" << endl;
    else cout << "NO" << endl;
}

signed main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr), cout.tie(nullptr);
    
    int T = 1;
    cin >> T;
    while (T--) solve();
    
    return 0;
}