F - Wormholes POJ - 3259（spfa判负环）

最新推荐文章于 2019-09-20 18:23:32 发布

Levi——

最新推荐文章于 2019-09-20 18:23:32 发布

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分类专栏：最短路

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本文介绍了一种使用SPFA算法解决时间旅行问题的方法，具体是在给定的农场地图中寻找是否存在一条路径，使得通过特定的虫洞可以回到出发点的时间点之前。

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Wormholes

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 60400		Accepted: 22564

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

题意：John的农场里N块地，M条路连接两块地（无向）没走过一条路，消耗w秒，V个虫洞（有向），走过虫洞，时间会倒退w秒。判段，是否可以以N块地中的一块地为起点，找到一条路，使得时间可以不断的倒退。

思路：spfa判断是否有负环！

#include "iostream"
#include "queue"
#include "vector"
#include "cstring"
using namespace std;
const int Max=1e3+10;
int N,M,V;
int vis[Max],d[Max],cnt[Max];//分别记录是否已经入队，时间消耗，和入队次数
struct  edge//记录目的地和话费，如果是虫洞的话，花费为负数
{
    int to,cost;
};
vector<edge> G[Max];
bool spfa(int s)
{
    vis[s]=1;
    d[s]=0;
    queue<int> que;
    que.push(s);
    while(!que.empty()){
        int u=que.front();
        que.pop();
        vis[u]=0;
        for(int i=0;i<G[u].size();i++){
            edge e=G[u][i];
            if(d[e.to]>d[u]+e.cost){
                d[e.to]=d[u]+e.cost;
                if(vis[e.to]) continue;
                que.push(e.to);
                vis[e.to]=1;
                if(++cnt[e.to]>N&&d[e.to]<0) return 1;//一个最多可以进队N次，如果大于N说明有环，如果时间为负数，说明是负环
            }
        }
    }
    return 0;
}
int main()
{
    ios::sync_with_stdio(false);
    int t;
    cin>>t;
    while(t--){
        for(int i=0;i<Max;i++){G[i].clear(); vis[i]=0; d[i]=0xfffffff;cnt[i]=0;}
        int u,v,w;
        edge e;
        cin>>N>>M>>V;
        for(int i=0;i<M;i++){//普通道路
            cin>>u>>v>>w;
            e.to=v,e.cost=w;
            G[u].push_back(e);
            e.to=u;
            G[v].push_back(e);
        }
        for(int i=0;i<V;i++){//虫洞
            cin>>u>>v>>w;
            e.to=v;
            e.cost=-w;
            G[u].push_back(e);
        }
        int flag=0;
        for(int i=1;i<=N;i++)
            if(spfa(i)){flag=1;break;}
        if(flag)
            cout<<"YES"<<endl;
        else
            cout<<"NO"<<endl;
    }
}