poj3259 Wormholes【Bellman-Ford或 SPFA判断是否有负环 】

题目链接：poj3259 Wormholes

题意：虫洞问题，有n个点，m条边为双向，还有w个虫洞（虫洞为单向，并且通过时间为倒流，即为负数），问你从任意某点走，能否穿越到之前。

贴个SPFA代码：

 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<iostream>
 #include<string>
 #include<stack>
 #include<queue>
 #include<vector>
 #include<set>
 #include<map>
 #include<cmath>
 using namespace std;
 #define lson rt*2
 #define rson rt*2+1
 #define CLR(a, b) memset((a), (b), sizeof((a)))
 typedef long long ll;
 typedef unsigned long long ull;
 typedef pair<int, int>pii;
 const double inf = 0x3f3f3f3f;
 const int N = ;
 const int M = *+;
 struct node {
     int u, v;
     int t;
     int nex;
 }edge[M];
 int head[N];

 int d[N];
 int vis[N];

 int n, m, w;

 int spfa(int x) {
     int i, j, k;
     queue<int> q;
     d[x] = ;
     vis[x] = ;
     q.push(x);
     while(!q.empty()) {
         int u = q.front(); q.pop();
         vis[u] = ;
         for(i = head[u]; ~i; i = edge[i].nex) {
             if(d[edge[i].v] > d[u] + edge[i].t) {
                 d[edge[i].v] = d[u] + edge[i].t;
                 if(!vis[edge[i].v]) {
                     vis[edge[i].v] = ;
                     q.push(edge[i].v);
                 }
             }
         }
         if(d[x] < ) return ;//存在负权环
     }
     return ;
 }
 int main() {
     int t, i;
     scanf("%d", &t);
     while(t--) {
         scanf("%d%d%d", &n, &m, &w);

         for(i = ; i <= n; ++i) head[i] = -;

         for(i = ; i <= *m; i += ) { //普通路 双向
             scanf("%d%d%d", &edge[i].u, &edge[i].v, &edge[i].t);
             edge[i].nex = head[edge[i].u];
             head[edge[i].u] = i;

             edge[i+].u = edge[i].v;
             edge[i+].v = edge[i].u;
             edge[i+].t = edge[i].t;
             edge[i+].nex = head[edge[i].v];
             head[edge[i].v] = i+;
         }
         for(; i <= *m+w; ++i) {//虫洞 单向
             scanf("%d%d%d", &edge[i].u, &edge[i].v, &edge[i].t);
             edge[i].t = -edge[i].t;
             edge[i].nex = head[edge[i].u];
             head[edge[i].u] = i;
         }

         for(i = ; i <= n; ++i) {
             d[i] = inf;
             vis[i] = ;
         }
         int f = ;
         for(i = ; i <= n; ++i) {
             if(d[i] == inf) {//图可能不连通
                 if(spfa(i)) {f = ; break;}
             }
         }
         if(f) puts("YES");
         else puts("NO");
     }
     return ;
 }

63ms