POJ 3259 Wormholes（最短路，判断有没有负环回路）

Wormholes

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 24249		Accepted: 8652

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

 

 

 

 

这题就是判断存不存在负环回路。

前M条是双向边，后面的W是单向的负边。

 

为了防止出现不连通，

增加一个结点作为起点。

起点到所有点的长度为0

 

bellman_ford算法：

/*

 * POJ 3259

 * 判断图中是否存在负环回路。

 * 为了防止图不连通的情况，增加一个点作为起点，这个点和其余的点都相连。

 */



#include <iostream>

#include <stdio.h>

#include <string.h>

#include <algorithm>

#include <vector>

#include <queue>

using namespace std;

/*

 * 单源最短路bellman_ford算法，复杂度O(VE)

 * 可以处理负边权图。

 * 可以判断是否存在负环回路。返回true,当且仅当图中不包含从源点可达的负权回路

 * vector<Edge>E;先E.clear()初始化，然后加入所有边

 * 点的编号从1开始(从0开始简单修改就可以了)

 */

const int INF=0x3f3f3f3f;

const int MAXN=550;

int dist[MAXN];

struct Edge

{

    int u,v;

    int cost;

    Edge(int _u=0,int _v=0,int _cost=0):u(_u),v(_v),cost(_cost){}

};

vector<Edge>E;

bool bellman_ford(int start,int n)//点的编号从1开始

{

    for(int i=1;i<=n;i++)dist[i]=INF;

    dist[start]=0;

    for(int i=1;i<n;i++)//最多做n-1次

    {

        bool flag=false;

        for(int j=0;j<E.size();j++)

        {

            int u=E[j].u;

            int v=E[j].v;

            int cost=E[j].cost;

            if(dist[v]>dist[u]+cost)

            {

                dist[v]=dist[u]+cost;

                flag=true;

            }

        }

        if(!flag)return true;//没有负环回路

    }

    for(int j=0;j<E.size();j++)

        if(dist[E[j].v]>dist[E[j].u]+E[j].cost)

            return false;//有负环回路

    return true;//没有负环回路

}



int main()

{

//    freopen("in.txt","r",stdin);

//    freopen("out.txt","w",stdout);

    int T;

    int N,M,W;

    int a,b,c;

    scanf("%d",&T);

    while(T--)

    {

        scanf("%d%d%d",&N,&M,&W);

        E.clear();

        while(M--)

        {

            scanf("%d%d%d",&a,&b,&c);

            E.push_back(Edge(a,b,c));

            E.push_back(Edge(b,a,c));

        }

        while(W--)

        {

            scanf("%d%d%d",&a,&b,&c);

            E.push_back(Edge(a,b,-c));

        }

        for(int i=1;i<=N;i++)

            E.push_back(Edge(N+1,i,0));

        if(!bellman_ford(N+1,N+1))printf("YES\n");

        else printf("NO\n");

    }

    return 0;

}

 

 

 

SPFA算法：

//============================================================================

// Name        : POJ.cpp

// Author      : 

// Version     :

// Copyright   : Your copyright notice

// Description : Hello World in C++, Ansi-style

//============================================================================



#include <iostream>

#include <string.h>

#include <stdio.h>

#include <algorithm>

#include <vector>

#include <queue>

using namespace std;

/*

 * 单源最短路SPFA

 * 时间复杂度 0(kE)

 * 这个是队列实现，有时候改成栈实现会更加快，很容易修改

 * 这个复杂度是不定的

 */

const int MAXN=1010;

const int INF=0x3f3f3f3f;

struct Edge

{

    int v;

    int cost;

    Edge(int _v=0,int _cost=0):v(_v),cost(_cost){}

};

vector<Edge>E[MAXN];

void addedge(int u,int v,int w)

{

    E[u].push_back(Edge(v,w));

}

bool vis[MAXN];

int cnt[MAXN];

int dist[MAXN];

bool SPFA(int start,int n)

{

    memset(vis,false,sizeof(vis));

    for(int i=1;i<=n;i++)dist[i]=INF;

    dist[start]=0;

    vis[start]=true;

    queue<int>que;

    while(!que.empty())que.pop();

    que.push(start);

    memset(cnt,0,sizeof(cnt));

    cnt[start]=1;

    while(!que.empty())

    {

        int u=que.front();

        que.pop();

        vis[u]=false;

        for(int i=0;i<E[u].size();i++)

        {

            int v=E[u][i].v;

            if(dist[v]>dist[u]+E[u][i].cost)

            {

                dist[v]=dist[u]+E[u][i].cost;

                if(!vis[v])

                {

                    vis[v]=true;

                    que.push(v);

                    if(++cnt[v]>n)return false;

                    //有负环回路

                }

            }

        }

    }

    return true;

}

int main()

{

//    freopen("in.txt","r",stdin);

//    freopen("out.txt","w",stdout);

    int T;

    int N,M,W;

    int a,b,c;

    scanf("%d",&T);

    while(T--)

    {

        scanf("%d%d%d",&N,&M,&W);

        for(int i=1;i<=N+1;i++)E[i].clear();

        while(M--)

        {

            scanf("%d%d%d",&a,&b,&c);

            addedge(a,b,c);

            addedge(b,a,c);

        }

        while(W--)

        {

            scanf("%d%d%d",&a,&b,&c);

            addedge(a,b,-c);

        }

        for(int i=1;i<=N;i++)

            addedge(N+1,i,0);

        if(!SPFA(N+1,N+1))printf("YES\n");

        else printf("NO\n");

    }

    return 0;

}