poj 3259 Wormholes
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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
23 3 11 2 21 3 42 3 13 1 33 2 11 2 32 3 43 1 8
Sample Output
NOYES
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.
Bellman-ford算法
题意:John的农场里N块地,M条路连接两块地(双向的),W个虫洞。虫洞是一条单向路,不但会把你传送到目的地,而且时间会倒退Ts(也就是负权边,所以我们存的是负值)。我们的任务是知道会不会在从某块地出发后又回来,看到了离开之前的自己。
思路:由于存在负权边,Dijkstra便不能用了,所以用bellman_ford。题目简化下,就是看图中有没有负权环,有的话John可以无限次走这个环,使得时间一定能得到一个负值。所以有的存在负环话就是可以,没有的话就是不可以了。
#include<iostream>
using namespace std;
struct node
{
int u,v,val;
}p[10000];
int N,M,W,k;
int dist[501],INF=1000000;
void addedge(int u,int v,int t)
{
p[k].u=u; p[k].v=v; p[k].val=t;
k++;
}
bool Bellman()
{
for(int i=1;i<=N;i++)
dist[i]=INF;
dist[1]=0;
for(int i=1;i<=N;i++) //边的松弛
for(int j=1;j<k;j++)
if(dist[p[j].v]>dist[p[j].u]+p[j].val)
dist[p[j].v]=dist[p[j].u]+p[j].val;
for(int i=1;i<k;i++) //查看是否存在负权边
if(dist[p[i].v]>dist[p[i].u]+p[i].val)
{
return true;
}
return false;
}
int main()
{
int T;
cin>>T;
while(T--)
{
int i,s,e,t;
cin>>N>>M>>W;
k=1;
for(i=0;i<M;i++)
{
cin>>s>>e>>t;
addedge(s,e,t);//农场间是双向的
addedge(e,s,t);
}
for(i=0;i<W;i++)
{
cin>>s>>e>>t;
addedge(e,s,-t);//虫洞是单向的
}
if(Bellman())
cout<<"YES"<<endl;
else
cout<<"NO"<<endl;
}
return 0;
}
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