题目链接：http://poj.org/problem?id=3259
Wormholes Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 44090 Accepted: 16203Description
While exploring his many farms, Farmer John has discovered a number of amazing wormholes. A wormhole is very peculiar because it is a one-way path that delivers you to its destination at a time that is BEFORE you entered the wormhole! Each of FJ's farms comprises N (1 ≤ N ≤ 500) fields conveniently numbered 1..N, M (1 ≤ M ≤ 2500) paths, and W (1 ≤ W ≤ 200) wormholes.
As FJ is an avid time-traveling fan, he wants to do the following: start at some field, travel through some paths and wormholes, and return to the starting field a time before his initial departure. Perhaps he will be able to meet himself :) .
To help FJ find out whether this is possible or not, he will supply you with complete maps to F (1 ≤ F ≤ 5) of his farms. No paths will take longer than 10,000 seconds to travel and no wormhole can bring FJ back in time by more than 10,000 seconds.
Input
Line 1: A single integer, F. F farm descriptions follow. Line 1 of each farm: Three space-separated integers respectively: N, M, and W Lines 2.. M+1 of each farm: Three space-separated numbers ( S, E, T) that describe, respectively: a bidirectional path between S and E that requires T seconds to traverse. Two fields might be connected by more than one path. Lines M+2.. M+ W+1 of each farm: Three space-separated numbers ( S, E, T) that describe, respectively: A one way path from S to E that also moves the traveler back T seconds.Output
Lines 1.. F: For each farm, output "YES" if FJ can achieve his goal, otherwise output "NO" (do not include the quotes).Sample Input
2 3 3 1 1 2 2 1 3 4 2 3 1 3 1 3 3 2 1 1 2 3 2 3 4 3 1 8Sample Output
NO YESHint
For farm 1, FJ cannot travel back in time. For farm 2, FJ could travel back in time by the cycle 1->2->3->1, arriving back at his starting location 1 second before he leaves. He could start from anywhere on the cycle to accomplish this. 题意：2个案例，3个点，3条路，1个虫洞，路是双向的，虫洞是单向的，单向的路为负，起点为1，看能不能穿越。 分析：SPFA判负环。 没有看模板，直接上代码，结果发现不知道WA了多少次，实在搞不下去了，最后，佳鑫大神帮我Debug了，我在建图的时候head初始化了，结果在SPFA里面有初始化为-1了，真是害死我了，最后还是RE了一次，数组开小了。不管怎么样，Dijkstra，SPFA，Floyd，Kruskal都可以不用看模板了，很开心。 #include <stdio.h> #include<iostream> #include <queue> #include <string.h> using namespace std; #define INF 0x3f3f3f3f int n,m,t; struct Edge { int v; int w; int next; } edge[300000]; int NE = 0; int dis[5500]; int head[5500]; bool vis[5500]; int cnt[5500]; void add(int u,int v,int d) { edge[NE].v = v; edge[NE].w = d; edge[NE].next = head[u]; head[u]= NE++; } bool SPFA() { for(int i=1; i<=n; i++) { dis[i] = INF; vis[i] = false; cnt[i] = 0; } dis[1] = 0; vis[1] = true; cnt[1] = 1; queue<int> Q; Q.push(1); while(!Q.empty()) { int u=Q.front(); Q.pop(); vis[u]=false; for(int i=head[u]; i!=-1; i=edge[i].next) { if(dis[edge[i].v]>dis[u]+edge[i].w) { dis[edge[i].v] = dis[u] + edge[i].w; if(!vis[edge[i].v]) { vis[edge[i].v] = true; Q.push(edge[i].v); if(++cnt[edge[i].v]>n) return true; } } } } return false; } int main() { //freopen("input.txt","r",stdin); int cases; scanf("%d",&cases); while(cases--) { NE = 0; memset(head,-1,sizeof(head)); scanf("%d%d%d",&n,&m,&t); for(int i=0; i<m; i++) { int a,b,d; scanf("%d%d%d",&a,&b,&d); add(a,b,d); add(b,a,d); } for(int i=0; i<t; i++) { int a,b,d; scanf("%d%d%d",&a,&b,&d); add(a,b,-d); } if(SPFA()) printf("YES\n"); else printf("NO\n"); } return 0; }
转载于:https://www.cnblogs.com/TreeDream/p/5749447.html
