AcWing 852. spfa判断负环

#include <bits/stdc++.h>
using namespace std;

const int N = 2005, M = 10005;

int head[N], cnt;
int n, m;
int dis[N], backup[N];
int step[N];  //到当前点最短路径跳转的步数
bool st[N];

struct
{
   
    int v, w;
    int nxt;
} edges[M];

int spfa()
{
   
    memset(dis, 0x3f, sizeof dis);
    memset(st, 0, sizeof st);
    memset(step, 0, sizeof step);
    queue<int> q;

    for (int i = 1; i <= n; ++i) {
   
        dis[i] = 0;
        q.push(i);
        st[i] = true;
        step[i] = 1;
    }
    //设置一个到所有点距离都为0的超级源点，若不存在负环，则求所有点的最短路时最多经过的步数不超过n步

    while (!q.empty())
    {
   
        int f = q.front();
        q.pop();
        st[f] = false;

        for(int e = head[f]; e != 0; e = edges[e].nxt) {
   
            int u = f, v = edges[e].v, w = edges[e].w;

            if (dis[v] > dis[u] + w) {
   
                dis[v] = dis[u] + w;
                step[v] = step[u] + 1;
                if (!st[v]) {
   
                    q.push(v);
                    st[v] = true;
                }
                if (step[v] > n)
                    return 1;   
            }
        }
    }

    return 0;
}

int main()
{
   
#ifndef ONLINE_JUDGE
    freopen("D:/VS CODE/C++/in.txt", "r", stdin);
    freopen("D:/VS CODE/C++/out.txt", "w", stdout);
#endif
    cin >> n >> m;

    for (int i = 0; i < m; ++i)
    {
   
        int u, v, w;
        scanf("%d %d %d", &u, &v, &w);
        edges[++cnt].nxt = head[u];
        edges[cnt].v = v;
        edges[cnt].w = w;
        head[u] = cnt;
    }

    if (spfa())
        printf("Yes");
    else 
        printf("No");

    fclose(stdin);
    fclose(stdout);
    return 0;
}