poj——3259——Wormholes

最新推荐文章于 2019-08-20 19:01:44 发布

心向阳光2014

最新推荐文章于 2019-08-20 19:01:44 发布

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探讨了在一个包含多个地点、路径及特殊虫洞的环境中，是否存在一种路径组合可以让旅行者回到出发前的时间点，通过贝尔曼-福特算法检测负权环来判断是否可行。

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Wormholes

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 43434		Accepted: 15955

题意：有n个点，m条无向边，w条虫洞（可以回退到进洞前），问是否可以利虫洞见到以前的自己。

思路：找负全环；

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

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

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <queue>
#include <stack>
#include <algorithm>
using namespace std;
struct node
{
    int u,v,w;
}que[10010];
int k;
int dis[1010];
  int t,n,m1,m2;
const int INF=0x3f3f3f3f;
void bell_man()
{
    for(int i=1;i<=n;i++)
        dis[i]=INF;
    dis[1]=0;
    int u,v,w;
    int mark;
    for(int i=0;i<n;i++)  //多跑一次，存在松弛？ 有负权环？：无；
    {
        mark=1;
        for(int j=0;j<k;j++)
        {
            u=que[j].u,v=que[j].v;
            w=que[j].w;
            if(dis[u]+w<dis[v])
            {
                mark=0;
                dis[v]=dis[u]+w;
            }
        }
        if(mark)
            break;
    }
    if(mark)
    printf("NO\n");
    else
        printf("YES\n");
}
int main()
{

    scanf("%d",&t);
    int u,v,w;
    while(t--)
    {
        scanf("%d%d%d",&n,&m1,&m2);
         k=0;
         for(int i=0;i<m1;i++)
         {
             scanf("%d%d%d",&u,&v,&w);
             que[k].u=u;que[k].v=v,que[k++].w=w;
             que[k].u=v;que[k].v=u,que[k++].w=w;
         }
         for(int i=0;i<m2;i++)
         {
             scanf("%d%d%d",&que[k].u,&que[k].v,&que[k].w);
             que[k].w=-que[k].w;;
             k++;
         }
         bell_man();
    }
    return 0;
}