poj 3259 Wormholes 判断负权值回路

Wormholes

Time Limit: 2000 MS Memory Limit: 65536 KB

64-bit integer IO format: %I64d , %I64u   Java class name: Main

[Submit] [Status] [Discuss]

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.

//Memory Time
//308K   204MS

#include<iostream>
#include <string.h>
using namespace std;

int dis[1001];      //源点到各点权值
const int max_w=10001;      //无穷远

struct weight
{
    int s;
    int e;
    int t;
}edge[5200];

int N,M,W_h; //N (1≤N≤500)fields 顶点数
             //M (1≤M≤2500)paths 正权双向边
             //W_h (1≤W≤200) wormholes 虫洞（回溯），负权单向边
int all_e;   //边集（边总数）

bool bellman()
{
    bool flag;

    /*relax*/

    for(int i=0;i<N-1;i++)  ///dis松弛的次数
    {
        flag=false;
        for(int j=0;j<all_e;j++) ///所有边集
            if(dis[edge[j].e] > dis[edge[j].s] + edge[j].t)  ///可以松弛  更新
            {
                dis[edge[j].e] = dis[edge[j].s] + edge[j].t;
                flag=true;         //relax对路径有更新
            }
        if(!flag)
            break;  //只要某一次relax没有更新，说明最短路径已经查找完毕，或者部分点不可达，可以跳出relax
    }///已经更新完毕了 所有边集都是两点之间的最短路

    /*Search Negative Circle*/

    for(int k=0;k<all_e;k++) ///遍历所有边集 如果还出现能够再次更新成最短的边  就表明出现负权值回路了
        if( dis[edge[k].e] > dis[edge[k].s] + edge[k].t)
            return true;
    return false;
}
int main(void)
{
    int u,v,w;

    int F;
    cin>>F;
    while(F--)
    {
        memset(dis,max_w,sizeof(dis));    //源点到各点的初始值为无穷，即默认不连通

        cin>>N>>M>>W_h;

        all_e=0;      //初始化指针

        /*read in Positive Paths*/

        for(int i=1;i<=M;i++)
        {
            cin>>u>>v>>w;
            edge[all_e].s=edge[all_e+1].e=u;
            edge[all_e].e=edge[all_e+1].s=v;
            edge[all_e++].t=w;
            edge[all_e++].t=w;               //由于paths的双向性，两个方向权值相等，注意指针的移动
        }

        /*read in Negative Wormholds*/

        for(int j=1;j<=W_h;j++)
        {
            cin>>u>>v>>w;
            edge[all_e].s=u;
            edge[all_e].e=v;
            edge[all_e++].t=-w;     //注意权值为负
        }

        /*Bellman-Ford Algorithm*/

        if(bellman())
            cout<<"YES"<<endl;
        else
            cout<<"NO"<<endl;
    }
    return 0;
}
//Memory Time
//308K   204MS

#include<iostream>
#include <string.h>
using namespace std;

int dis[1001];      //源点到各点权值
const int max_w=10001;      //无穷远

struct weight
{
    int s;
    int e;
    int t;
}edge[5200];

int N,M,W_h; //N (1≤N≤500)fields 顶点数
             //M (1≤M≤2500)paths 正权双向边
             //W_h (1≤W≤200) wormholes 虫洞（回溯），负权单向边
int all_e;   //边集（边总数）

bool bellman()
{
    bool flag;

    /*relax*/

    for(int i=0;i<N-1;i++)  ///dis松弛的次数
    {
        flag=false;
        for(int j=0;j<all_e;j++) ///所有边集
            if(dis[edge[j].e] > dis[edge[j].s] + edge[j].t)  ///可以松弛  更新
            {
                dis[edge[j].e] = dis[edge[j].s] + edge[j].t;
                flag=true;         //relax对路径有更新
            }
        if(!flag)
            break;  //只要某一次relax没有更新，说明最短路径已经查找完毕，或者部分点不可达，可以跳出relax
    }///已经更新完毕了 所有边集都是两点之间的最短路

    /*Search Negative Circle*/

    for(int k=0;k<all_e;k++) ///遍历所有边集 如果还出现能够再次更新成最短的边  就表明出现负权值回路了
        if( dis[edge[k].e] > dis[edge[k].s] + edge[k].t)
            return true;
    return false;
}
int main(void)
{
    int u,v,w;

    int F;
    cin>>F;
    while(F--)
    {
        memset(dis,max_w,sizeof(dis));    //源点到各点的初始值为无穷，即默认不连通

        cin>>N>>M>>W_h;

        all_e=0;      //初始化指针

        /*read in Positive Paths*/

        for(int i=1;i<=M;i++)
        {
            cin>>u>>v>>w;
            edge[all_e].s=edge[all_e+1].e=u;
            edge[all_e].e=edge[all_e+1].s=v;
            edge[all_e++].t=w;
            edge[all_e++].t=w;               //由于paths的双向性，两个方向权值相等，注意指针的移动
        }

        /*read in Negative Wormholds*/

        for(int j=1;j<=W_h;j++)
        {
            cin>>u>>v>>w;
            edge[all_e].s=u;
            edge[all_e].e=v;
            edge[all_e++].t=-w;     //注意权值为负
        }

        /*Bellman-Ford Algorithm*/

        if(bellman())
            cout<<"YES"<<endl;
        else
            cout<<"NO"<<endl;
    }
    return 0;
}