POJ 3259 Wormholes（最短路径，求负环）

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

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 8

Sample Output

NO YES

Http

POJ:https://vjudge.net/problem/POJ-3259

Source

最短路径，求负环

题目大意

给出若干双向道路和单向虫洞，虫洞可以回复时间（即边权为负），而正常的双向道路需要花费一定的时间，现在求能否通过虫洞求得一个负环

解决思路

运用spfa算法，对每一个点做一个统计，统计入队的次数，如果发现次数>=点数，则说明有负环

代码

#include<iostream> #include<cstdio> #include<cstdlib> #include<algorithm> #include<cstring> #include<vector> #include<queue> using namespace std; const int maxN=600; const int inf=2147483647; class Edge { public: int v,w; }; int n,m,m2; vector<Edge> E[maxN]; bool inqueue[maxN]; queue<int> Q; int Dist[maxN]; int Tot[maxN]; int main() { int T; while (cin>>T) { for (int ti=1;ti<=T;ti++) { scanf("%d%d%d",&n,&m,&m2); for (int i=1;i<=n;i++) { Dist[i]=inf; E[i].clear(); } for (int i=1;i<=m;i++) { int u,w,v; scanf("%d%d%d",&u,&v,&w); E[u].push_back((Edge){v,w}); E[v].push_back((Edge){u,w}); } for (int i=1;i<=m2;i++) { int u,v,w; scanf("%d%d%d",&u,&v,&w); E[u].push_back((Edge){v,-w}); } memset(inqueue,0,sizeof(inqueue)); memset(Tot,0,sizeof(Tot)); while (!Q.empty()) Q.pop(); Dist[1]=0; Q.push(1); inqueue[1]=1; bool get=0; do { int u=Q.front(); Q.pop(); inqueue[u]=0; Tot[u]++; if (Tot[u]>=n) { get=1; break; } for (int i=0;i<E[u].size();i++) { int v=E[u][i].v; int w=E[u][i].w; if (Dist[v]>Dist[u]+w) { Dist[v]=Dist[u]+w; if (inqueue[v]==0) { Q.push(v); inqueue[v]=1; } } } } while (!Q.empty()); if (get) cout<<"YES"<<endl; else cout<<"NO"<<endl; } } return 0; }

转载于:https://www.cnblogs.com/SYCstudio/p/7226467.html

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