负环

虫洞

\(本题为裸题\)

求负环

\(基于spfa\)

• \((1)统计每个点入队的次数,如果某个点入队n次,则说明存在负环.\)
• \((2)统计每个点的最短路中所包含的边数,如果某点的最短路所包含的边数大于等于n,\\则也说明存在负环.\\\)
  \(对于求负环是否存在,dist的初始化是无所谓的.\)

#include <bits/stdc++.h>
using namespace std;
#define IO ios::sync_with_stdio(false);cin.tie(0); cout.tie(0)
inline int lowbit(int x) { return x & (-x); }
#define ll long long
#define pb push_back
#define PII pair<int, int>
#define x first
#define y second
#define inf 0x3f3f3f3f
const int N = 510, M = 5210;
int n, m1, m2;
int h[N], e[M], w[M], ne[M], idx;
int dist[N];
int q[N], cnt[N];
bool st[N];

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

bool spfa() {
    memset(cnt, 0, sizeof cnt);
    memset(st, 0, sizeof st);
    int hh = 0, tt = 0;
    for (int i = 1; i <= n; ++i) {
        q[tt++] = i;
        st[i] = 1;
    }
    while (hh != tt) {
        int t = q[hh++];
        if (hh == N) hh = 0;
        st[t] = false;
        for (int i = h[t]; ~i; i = ne[i]) {
            int j = e[i];
            if (dist[j] > dist[t] + w[i]) {
                dist[j] = dist[t] + w[i];
                cnt[j] += 1;
                if (cnt[j] >= n) return true;
                if (!st[j]) {
                    q[tt++] = j;
                    if (tt == N) tt = 0;
                    st[j] = true;
                }
            }
        }
    }
    return false;
}

int main() {
    IO;
    int _;
    cin >> _;
    while (_--) {
        cin >> n >> m1 >> m2;
        memset(h, -1, sizeof h);
        idx = 0;
        for (int i = 0; i < m1; ++i) {
            int a, b, c;
            cin >> a >> b >> c;
            add(a, b, c), add(b, a, c);
        }
        for (int i = 0; i < m2; ++i) {
            int a, b, c;
            cin >> a >> b >> c;
            add(a, b, -c);
        }
        if (spfa()) puts("YES");
        else puts("NO");
    }
    return 0;
}