Luo Gu problem solution [P2850] [USACO06DEC] Wormhole Wormholes

Face questions

SPFA forfeit ring bare title.

Analyzing the main ring SPFA negative \ (2 \) methods:

1. Determining whether each point is equal to the number of dequeued greater than \ (n-\) ;
2. Determining whether the number of points on each point of greater than or equal shortest path \ (n-\) .

Both methods are well documented. I personally recommend using the second method.

So, we just record it on a number of points for each point on the shortest path in the course of the SPFA.

Because we started all the points into the queue, so the \ (dist \) array must be initialized to \ (0 \) .

#include <bits/stdc++.h>

using namespace std;

const int N = 503, M = 5203;

int n, m, f, w;
int tot, head[N], ver[M], edge[M], nxt[M];
int dist[N], cnt[N];
bool st[N];

inline void add(int u, int v, int w)
{
    ver[++tot] = v, edge[tot] = w, nxt[tot] = head[u], head[u] = tot;
}

inline bool SPFA()
{
    memset(dist, 0, sizeof dist);
    memset(cnt, 0, sizeof cnt);
    memset(st, false, sizeof st);
    queue <int> q;
    for (int i = 1; i <= n; i+=1)
    {
        q.push(i); //一开始先将所有的点加入队列
        st[i] = true;
    }
    while (!q.empty())
    {
        int u = q.front(); q.pop();
        st[u] = false;
        for (int i = head[u]; i; i = nxt[i])
        {
            int v = ver[i], w = edge[i];
            if (dist[v] > dist[u] + w)
            {
                dist[v] = dist[u] + w;
                cnt[v] = cnt[u] + 1; //记录每个点最短路经的条数
                if (cnt[v] >= n) return true; //存在负环
                if (!st[v])
                {
                    st[v] = true;
                    q.push(v);
                }
            }
        }
    }
    return false;
}

int main()
{
    cin >> f;
    while (f--)
    {
        cin >> n >> m >> w;
        memset(head, 0, sizeof head);
        tot = 0;
        for (int i = 1; i <= m; i+=1)
        {
            int u, v, w;
            cin >> u >> v >> w;
            add(u, v, w), add(v, u, w);
        }
        for (int i = 1; i <= w; i+=1)
        {
            int u, v, w;
            cin >> u >> v >> w;
            add(u, v, -w);
        }
        if (SPFA()) puts("YES");
        else puts("NO");
    }
    return 0;
}