POJ 3259:Wormholes bellman_ford判定负环
Time Limit: 2000MS | Memory Limit: 65536K | |
Total Submissions: 37906 | Accepted: 13954 |
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.
正常的path是双向的,有一定的消耗时间。虫洞是单向的,能够让时间倒流一定时间。问FJ能否找到一条路径,能够遇见之前的那个自己。
说白了就是找负环。
Bellman_ford模板题,用来对每一条边都进行松弛,然后看最后结果是否依然能够松弛。如果还能松弛,说明有负环;如果不能松弛了,就是没有负环。
代码:
#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
#include <string>
#include <cstring>
#pragma warning(disable:4996)
using namespace std; struct E{
int s;
int e;
int l;
}edge[5205]; int N,M,W,edge_num;
int dis[505]; void addedge(int start,int end,int len)
{
edge_num++; edge[edge_num].s=start;
edge[edge_num].e=end;
edge[edge_num].l=len;
} bool bellman_ford()
{
int i,j;
for(i=1;i<=N-1;i++)
{
int flag=0;
for(j=1;j<=edge_num;j++)
{
if(dis[edge[j].e]>dis[edge[j].s]+edge[j].l)
{
flag=1;
dis[edge[j].e]=dis[edge[j].s]+edge[j].l;
}
}
if(flag==0)
break;
}
for(j=1;j<=edge_num;j++)
{
if(dis[edge[j].e]>dis[edge[j].s]+edge[j].l)
{
return true;
}
}
return false;
} int main()
{
int i,start,end,len;
int Test;
cin>>Test; while(Test--)
{
edge_num=0;
memset(dis,0,sizeof(dis)); cin>>N>>M>>W; for(i=1;i<=M;i++)
{
cin>>start>>end>>len; addedge(start,end,len);
addedge(end,start,len);
}
for(i=1;i<=W;i++)
{
cin>>start>>end>>len; addedge(start,end,-len);
}
if(bellman_ford())
{
cout<<"YES"<<endl;
}
else
{
cout<<"NO"<<endl;
}
}
return 0;
}
版权声明:本文为博主原创文章,未经博主允许不得转载。
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