poj3259: Wormholes(BF模板题）
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 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.
题目大意：虫洞问题，现在有n个点，m条边，代表现在可以走的通路，比如从a到b和从b到a需要花费c时间，现在在地上出现了w个虫洞，虫洞的意义就是你从a到b话费的时间是-c(时间倒流,并且虫洞是单向的)，现在问你从某个点开始走，能回到从前
解题思路：其实给出了坐标，这个时候就可以构成一张图，然后将回到从前理解为是否会出现负权环，用bellman-ford就可以解出了
#include <stdio.h> #include <string.h> #include <stdlib.h> #define N 900001 struct node { int u,v,w; }q[100001]; int dis[10001]; int n,m,w1,count=0; int B() { int flag=0; for(int i=1;i<=n;i++) dis[i]=N; dis[1]=0; for(int i=1;i<=n-1;i++) { flag=0; for(int j=0;j<count;j++) { if(dis[q[j].v]>dis[q[j].u]+q[j].w)//这里u,v是不能颠倒的，因为 { dis[q[j].v]=dis[q[j].u]+q[j].w; flag=1; } } /* for(int j=0;j<n;j++) printf(".%d",dis[j]);*/ if(!flag) break; } for(int i=0;i<count;i++) { if(dis[q[i].v]>dis[q[i].u]+q[i].w) return 0; } return 1; } int main() { int x,y,x1,T; scanf("%d",&T); while(T--) { count=0; scanf("%d%d%d",&n,&m,&w1); while(m--) { scanf("%d%d%d",&x,&y,&x1); q[count].u=x; q[count].v=y; q[count++].w=x1; q[count].u=y; q[count].v=x; q[count++].w=x1; } while(w1--) { scanf("%d%d%d",&x,&y,&x1); q[count].u=x; q[count].v=y; q[count++].w=-x1;//他是有方向的 } int t=B(); if(t==0) printf("YES\n"); else printf("NO\n"); } return 0; }
第二次写的：
#include <iostream> #include <stdio.h> #include <string.h> #include <stdlib.h> #define INF 0x7fffffff using namespace std; int n,m,k,tt; struct node { int x,y,z; } q[10001]; int dis[1001]; void add(int xx,int yy,int zz) { q[tt].x=xx; q[tt].y=yy; q[tt++].z=zz; } void BF() { int flag; dis[1]=0; for(int i=1; i<=n; i++) { flag=0; for(int i=0; i<tt; i++) { if(dis[q[i].y]>dis[q[i].x]+q[i].z) { dis[q[i].y]=dis[q[i].x]+q[i].z; flag=1; } } if(flag==0) break; } if(flag==1) printf("YES\n"); else printf("NO\n"); } int main() { int T,zz,xx,yy; cin>>T; while(T--) { cin>>n>>m>>k; tt=0; for(int i=0; i<m; i++) { cin>>xx>>yy>>zz; add(xx,yy,zz); add(yy,xx,zz); } for(int i=0; i<k; i++) { cin>>xx>>yy>>zz; add(xx,yy,-zz); } BF(); } return 0; }
