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 8Sample Output
NO
YESHint
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.
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;
}View Code
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;
}View Code
