POJ – Wormholes-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 8

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

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花费的时间是-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”，只要判断是否有负权环就行了。

Bellman-Ford算法

• 解决含负权边的带权有向图的单源最短路径问题 不能处理带负权边的无向图(因可以来回走一条负权边)
• 限制条件： 要求图中不能包含权值总和为负值回路(负权值回路)，如下图所示。
  
• 最多n-1条边，最多n-1轮就可以实现最短路径，当第n轮路径还在改变，说明此时存在负值回路。

#include<stdio.h>
#include<math.h>
#include<vector>
using namespace std;

const int INF=1<<30;
struct Edge
{
    int s,e,w;
    Edge(int ss,int ee,int ww):s(ss),e(ee),w(ww){ }
    Edge() { }
};
vector<Edge> edges;//所有的边
int N,M,W;
int dist[1005];
int Bellman_ford(int v)
{
    for(int i=1;i<=N;++i)
        dist[i]=INF;
    dist[v]=0;
    for(int k=1;k<N;++k){//N-1轮松弛
        bool flag=false;//对于此题改进操作，若不需要更新了就直接返回假
        for(int i=0;i<edges.size();++i){//扫描所有边
            int s=edges[i].s;
            int e=edges[i].e;
            if(dist[s]+edges[i].w<dist[e]){//松弛
                dist[e]=dist[s]+edges[i].w;
                flag=true;
            }
        }
        if(!flag) return false;//提早结束，不会有负权环
    }
    //再进行1轮松弛，检查有没有负权边，如果出现更短路，则有负权边
    for(int i=0;i<edges.size();++i){
        int s=edges[i].s;
        int e=edges[i].e;
        if(dist[s]+edges[i].w<dist[e])
            return true;
    }
    return false;
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--){
        edges.clear();
        scanf("%d%d%d",&N,&M,&W);
        for(int i=0;i<M;++i){
            int s,e,t;
            scanf("%d%d%d",&s,&e,&t);
            edges.push_back(Edge(s,e,t));//双向边等于两条边
            edges.push_back(Edge(e,s,t));
        }
        for(int i=0;i<W;++i){
            int s,e,t;
            scanf("%d%d%d",&s,&e,&t);
            edges.push_back(Edge(s,e,-t));
        }
        if(Bellman_ford(1))//从1可达所有点
            printf("YES\n");
        else
            printf("NO\n");
    }
    return 0;
}

SPFA算法

• SPFA算法是Bellman-Ford算法的改进版，用队列维护dist[]。

#include<cstdio>
#include<vector>
#include<queue>
#include<cstring>
using namespace std;

const int INF=1<<30;
struct Edge
{
    int e,w;
    Edge(int ee,int ww):e(ee),w(ww){ }
    Edge() { }
};
vector<Edge> G[1005]; //整个有向图
int updateTimes[1005];//最短路的改进次数
int dist[1005];//dist[i]是源到i的目前最短路长度
int N,M,W;

bool Spfa(int v)
{
    for(int i=1;i<=N;++i)//dis初始化
        dist[i]=INF;
    dist[v]=0;//源点v
    queue<int> que;//定义队列que
    que.push(v);//源点v入队列que
    memset(updateTimes,0,sizeof(updateTimes));
    while(!que.empty()){
        int s=que.front();
        que.pop();
        for(int i=0;i<G[s].size();++i){
            int e=G[s][i].e;
            if(dist[e]>dist[s]+G[s][i].w){
                dist[e]=dist[s]+G[s][i].w;
                que.push(e);//没判队列里是否已经有e，可能会慢一些
                ++updateTimes[e];
                if(updateTimes[e]>=N) return true;
            }
        }
    }
    return false;
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--){
        scanf("%d%d%d",&N,&M,&W);
        for(int i=1;i<1000;++i) G[i].clear();
        int s,e,t;
        for(int i=0;i<M;++i){
            scanf("%d%d%d",&s,&e,&t);
            G[s].push_back(Edge(e,t));//双向边等于两条边
            G[e].push_back(Edge(s,t));
        }
        for(int i=0;i<W;++i){
            scanf("%d%d%d",&s,&e,&t);
            G[s].push_back(Edge(e,-t));
        }
        if(Spfa(1))//从1可达所有点
            printf("YES\n");
        else
            printf("NO\n");
    }
    return 0;
}