poj3259

Wormholes

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 24864		Accepted: 8869

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 toF (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

#include<iostream>

#include<stdio.h>

using namespace std;

const int fMax = 505;

const int eMax = 5205;

const int wMax = 99999;

struct{

    int sta, end, time;

}edge[eMax];

int point_num, edge_num, dict[fMax];

bool bellman_ford()

{

    int i, j;

    for(i = 2; i <= point_num; i ++)

        dict[i] = wMax;//初始化

    for(i = 1; i < point_num; i ++)//点要减1

	{

        bool finish = true;    //  加个全部完成松弛的判断，优化了50多MS。 

        for(j = 1; j <= edge_num; j ++)

		{

            int u = edge[j].sta;

            int v = edge[j].end;

            int w = edge[j].time;

            if(dict[v] > dict[u] + w)

			{   //  松弛。

                dict[v] = dict[u] + w;

                finish = false;

            }

        }

        if(finish)  break;

    }

    for(i = 1; i <= edge_num; i ++)

	{   //  是否存在负环的判断。

        int u = edge[i].sta;

        int v = edge[i].end;

        int w = edge[i].time;

        if(dict[v] > dict[u] + w) 

			

            return false;

    }

    return true;

}

int main()

{

    int farm;

    scanf("%d", &farm);

    while(farm --)

	{

        int field, path, hole;

        scanf("%d %d %d", &field, &path, &hole);

        int s, e, t, i, k = 0;

        for(i = 1; i <= path; i ++)

		{

            scanf("%d %d %d", &s, &e, &t);  //  用scanf代替了cin，优化了100多MS。

            k ++;

            edge[k].sta = s;

            edge[k].end = e;

            edge[k].time = t;

            k ++;

            edge[k].sta = e;

            edge[k].end = s;

            edge[k].time = t;

        }

        for(i = 1; i <= hole; i ++)

		{

            scanf("%d %d %d", &s, &e, &t);

            k ++;

            edge[k].sta = s;

            edge[k].end = e;

            edge[k].time = -t;

        }

        point_num = field;

        edge_num = k;

        if(!bellman_ford())  

			printf("YES\n");

        else  printf("NO\n");

		for(i=0;i<=point_num;i++)

			printf("%d  ",dict[i]);

    }

    return 0;

}