POJ 3259—Wormholes
阿新 • • 發佈:2018-11-25
Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 66356 Accepted: 24748 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 Hint 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. Source USACO 2006 December Gold
解題報告(註釋題解):
#include<iostream> using namespace std; const int inf=10010; struct node{ int f,t,c; }; int S,E,T,N,M,W,NUM; node n[5200]; int bellman_ford() { int dis[505]; dis[0]=0; for(int i=1;i<N;i++)dis[i]=inf;//開始時設每一個點到別的點的距離無限大。就是每個點的權值 for(int i=1;i<N;i++)//以這個圖的點得個數為外層迴圈,,由於是從任一點開始,所以這層迴圈求的是任一點的最短距離。 { int flag=0; for(int j=1;j<=NUM;j++)//以有向邊的個數為內層迴圈,對每一個邊進行鬆弛。 { if(dis[n[j].t]>dis[n[j].f]+n[j].c) { dis[n[j].t]=dis[n[j].f]+n[j].c; flag=1;//當迴圈進不來時說明不能鬆弛了。 } } if(!flag)break;//不能鬆弛就跳出迴圈。 } for(int i=1;i<=NUM;i++) { if(dis[n[i].t]>dis[n[i].f]+n[i].c) return true; } return false; } int main() { int nn; cin>>nn; while(nn--) { NUM=1; cin>>N>>M>>W; while(M--) { cin>>S>>E>>T; n[NUM].f=S; n[NUM].t=E; n[NUM++].c=T;//無向邊,故需要將兩個方向都記下。 n[NUM].f=E; n[NUM].t=S; n[NUM++].c=T; } while(W--) { cin>>S>>E>>T; n[NUM].f=S; n[NUM].t=E; n[NUM++].c=(0-T); } if(bellman_ford()) cout<<"YES"<<endl; else cout<<"NO"<<endl; } return 0; }