POJ3259:Wormholes(spfa判负环)
作者:互联网
Wormholes
Time Limit: 2000MS | Memory Limit: 65536K | |
Total Submissions: 68097 | Accepted: 25374 |
题目链接:http://poj.org/problem?id=3259
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
题意:
给出一些双向边以及其对应花费,然后还有一些单向边,可以回到之前的时间。问最后能否从起点出发,回到起点的时候时间在出发之前。
题解:
对于时间通道建负边权就行了,然后跑spfa看有没得负环。
代码如下:
#include <cstdio> #include <cstring> #include <algorithm> #include <iostream> #include <queue> #define INF 0x3f3f3f3f using namespace std; typedef long long ll; const int N = 505,M = 2505; int T,n,m,w; int head[N],d[N],c[N],vis[N]; struct Edge{ int u,v,w,next; }e[M<<2]; int tot; void adde(int u,int v,int w){ e[tot].u=u;e[tot].v=v;e[tot].w=w;e[tot].next=head[u];head[u]=tot++; } int spfa(int s){ queue <int> q; memset(d,INF,sizeof(d)); memset(vis,0,sizeof(vis));memset(c,0,sizeof(c)); q.push(s);vis[s]=1;d[s]=0;c[s]=1; while(!q.empty()){ int u=q.front();q.pop();vis[u]=0; if(c[u]>n){ return 1; } for(int i=head[u];i!=-1;i=e[i].next){ int v=e[i].v; if(d[v]>d[u]+e[i].w){ d[v]=d[u]+e[i].w; if(!vis[v]){ vis[v]=1; q.push(v); c[v]++; } } } } return 0; } int main(){ scanf("%d",&T); while(T--){ scanf("%d%d%d",&n,&m,&w); memset(head,-1,sizeof(head));tot=0; for(int i=1;i<=m;i++){ int u,v,c; scanf("%d%d%d",&u,&v,&c); adde(u,v,c);adde(v,u,c); } for(int i=1;i<=w;i++){ int u,v,c; scanf("%d%d%d",&u,&v,&c); adde(u,v,-c); } int flag=0; if(spfa(1)) flag=1; if(flag) puts("YES"); else puts("NO"); } return 0; }
标签:his,farm,vis,int,判负,spfa,FJ,include,POJ3259 来源: https://www.cnblogs.com/heyuhhh/p/10352102.html