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;
}