POJ 3259 Wormholes (Bellman)
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 23270 | Accepted: 8301 |
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 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
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
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.
Source
思路：用bellman-ford 判断有没有负权回路，如果有他就能看到自己。 不过，我认为应该判断每个点有没有负权回路，而不仅仅只判断第一个点就行了（如果某位大牛路过看到，觉得理解不对 希望多多指教）
#include<iostream> #include<cstdio> #include<cstring> using namespace std; const int VM=520; const int EM=2520; const int INF=0x3f3f3f3f; struct Edge{ int u,v; int cap; }edge[EM<<1]; int n,m,k; int cnt,dis[VM]; void addedge(int cu,int cv,int cw){ edge[cnt].u=cu; edge[cnt].v=cv; edge[cnt].cap=cw; cnt++; } int Bellman_ford(){ for(int i=1;i<=n;i++) dis[i]=INF; dis[1]=0; for(int i=1;i<n;i++) //n-1次松弛 for(int j=0;j<cnt;j++) //枚举每条边 if(dis[edge[j].v]>dis[edge[j].u]+edge[j].cap) dis[edge[j].v]=dis[edge[j].u]+edge[j].cap; for(int j=0;j<cnt;j++) if(dis[edge[j].v]>dis[edge[j].u]+edge[j].cap) //判断是否存在负权边 return 0; return 1; } int main(){ //freopen("input.txt","r",stdin); int t; scanf("%d",&t); while(t--){ scanf("%d%d%d",&n,&m,&k); cnt=0; int u,v,w; while(m--){ scanf("%d%d%d",&u,&v,&w); addedge(u,v,w); addedge(v,u,w); } while(k--){ scanf("%d%d%d",&u,&v,&w); addedge(u,v,-w); } int ans=Bellman_ford(); printf("%s\n",ans==0?"YES":"NO"); } return 0; }
下面这个代码稍快一点：
#include<iostream> #include<cstdio> #include<cstring> using namespace std; const int VM=520; const int EM=2520; const int INF=0x3f3f3f3f; struct node{ int u,v; int cap; }edge[EM<<1]; int n,m,k; int cnt,dis[VM]; void addedge(int cu,int cv,int cw){ edge[cnt].u=cu; edge[cnt].v=cv; edge[cnt].cap=cw; cnt++; } int Bellman_ford(){ int i,j; for(i=1;i<=n;i++) dis[i]=INF; dis[1]=0; for(i=1;i<=n;i++){ int flag=0; for(j=0;j<cnt;j++) if(dis[edge[j].v]>dis[edge[j].u]+edge[j].cap){ dis[edge[j].v]=dis[edge[j].u]+edge[j].cap; flag=1; } if(!flag) //优化 break; } return i==n+1; //相等则存在正环 } int main(){ //freopen("input.txt","r",stdin); int t; scanf("%d",&t); while(t--){ scanf("%d%d%d",&n,&m,&k); cnt=0; int u,v,w; while(m--){ scanf("%d%d%d",&u,&v,&w); addedge(u,v,w); addedge(v,u,w); } while(k--){ scanf("%d%d%d",&u,&v,&w); addedge(u,v,-w); } int ans=Bellman_ford(); printf("%s\n",ans==1?"YES":"NO"); } return 0; }
