POJ 3259 最短路 判负环

SPFA 判负环 某个点访问n次

#include <set>#include <map>#include <cmath>#include <queue>#include <stack>#include <string>#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>using namespace std;typedef  long long LL;const double PI = acos(-1.0);template <class T> inline  T MAX(T a, T b){if (a > b) return a;return b;}template <class T> inline  T MIN(T a, T b){if (a < b) return a;return b;}const int N = 111;const int M = 11111;const LL MOD = 1000000007LL;const int dir[4][2] = {1, 0, -1, 0, 0, -1, 0, 1};const int INF = 0x3f3f3f3f;int dist[555];struct node{    int v, w, next;}edge[6666];int head[555], cnt;int n, m, vis[555];int inq[555];inline void addnode(int u, int v, int w){    edge[cnt].v = v; edge[cnt].next = head[u];    edge[cnt].w = w; head[u] = cnt++;}bool solve(){    queue < int > q;    fill(dist, dist + n + 1, INF);    memset(vis, 0, sizeof(vis));    memset(inq, 0, sizeof(inq));    dist[1] = 0; vis[1]++; inq[1]++;    int u, i, j, k, v;    q.push(1);    while (!q.empty())    {        u = q.front(); q.pop(); inq[u]--;        for (i = head[u]; ~i; i = edge[i].next)        {            v = edge[i].v;            if (dist[u] + edge[i].w < dist[v])            {                dist[v] = dist[u] + edge[i].w;                if (!inq[v]) {q.push(v); vis[v]++; inq[v]++; if (vis[v] >= n) return true;}            }        }    }    return false;}int main(){    int T;    scanf("%d", &T);    while (T--)    {        int u, v, w, k, i;        memset(head, -1, sizeof(head));        cnt = 0;        scanf("%d%d%d", &n, &m, &k);        for (i = 0; i < m; ++i)        {            scanf("%d%d%d", &u, &v, &w);            addnode(u, v, w);            addnode(v, u, w);        }        for (i = 0; i < k; ++i)        {            scanf("%d%d%d", &u, &v, &w);            addnode(u, v, -w);        }        int flag = 0;        if (solve()) printf("YES\n");        else printf("NO\n");    }    return 0;}