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Wormholes Time Limit:2000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u
Description
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
Output
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 8
Sample Output
NO YES
Hint
#include<iostream> #include<vector> #include<queue> using namespace std; #define N 550 #define INF 0xffffff struct node { int y, w; }P[N]; vector<node> G[N]; int v[N]; int spfa(int s) { queue<int> Q; Q.push(s); while(Q.size()) { s = Q.front(); Q.pop(); int len = G[s].size(); for(int i = 0; i < len; i++) { node q = G[s][i]; if(v[s] + q.w < v[q.y]) { v[q.y] = v[s] + q.w; Q.push(q.y); } } if(v[1] < 0) return 1; } return 0; } int main() { int t, n, m, w, s, e, f; cin >> f; while(f--) { for(int i = 1; i < N; i++) { v[i] = INF; G[i].clear(); } v[1] = 0; cin >> n >> m >> w; node p; for(int i = 0; i < m; i++) { cin >> s >> e >> t; p.y = e, p.w = t; G[s].push_back(p); p.y = s; G[e].push_back(p); } for(int i = 0; i < w; i++) { cin >> s >> e >> t; p.y = e, p.w = -t; G[s].push_back(p); } int ans = spfa(1); if(ans) cout << "YES" << endl; else cout << "NO" << endl; } return 0; }
SPFA 在形式上和宽度优先搜索非常类似，不同的是宽度优先搜索中一个点出了队列就不可能重新进入队列，但是SPFA中一个点可能在出队列之后再次被放入队列，也就是一个点改进过其它的点之后，过了一段时间可能本身被改进，于是再次用来改进其它的点，这样反复迭代下去。
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原文地址：http://www.cnblogs.com/Tinamei/p/4678591.html
