poj-3259 Wormholes（无向、负权、最短路之负环判断）

http://poj.org/problem?id=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.

 

题目大意：

时空旅行，前m条路是双向的，旅行时间为正值，w条路是虫洞，单向的，旅行时间是负值，也就是能回到过去。求从一点出发，判断能否在”过去“回到出发点，即会到出发点的时间是负的。

解题思路：

裸的负权最短路问题，SPFA Bellman-Ford解决。

#include<iostream>
#include<cstdio>
using namespace std;
#define INF 0x3f3f3f3f
#define N 10100
int nodenum, edgenum, w, original=1; //点，边，起点

typedef struct Edge //边
{
    int u;
    int v; 
    int cost;
}Edge;//边的数据结构

Edge edge[N];//边

int dis[N];//距离

bool Bellman_Ford()
{
    for(int i = 1; i <= nodenum; ++i) //初始化
        dis[i] = (i == original ? 0 : INF);
    int F=0; 
    for(int i = 1; i <= nodenum - 1; ++i)//进行nodenum-1次的松弛遍历
    {
        for(int j = 1; j <= edgenum*2+w; ++j)
        {
            if(dis[edge[j].v] > dis[edge[j].u] + edge[j].cost) //松弛（顺序一定不能反~）
            {
                dis[edge[j].v] = dis[edge[j].u] + edge[j].cost;
                F=1;
            }
        }
        if(!F)
            break;
    }    
        //与迪杰斯特拉算法类似，但不是贪心！
        //并没有标记数组
        //本来松弛已经结束了
        //但是因为由于负权环的无限松弛性
        bool flag = 1; //判断是否含有负权回路
        //如果存在负权环的话一定能够继续松弛
        for(int i = 1; i <= edgenum*2+w; ++i)
        {
            if(dis[edge[i].v] > dis[edge[i].u] + edge[i].cost)
            {
                flag = 0;
                break;
            }
        }
        //只有在负权环中才能再松弛下去
    return flag;
}

int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        
        scanf("%d %d %d", &nodenum, &edgenum, &w);

        for(int i = 1; i <= 2*edgenum; i+=2)//加上道路，双向边 
        {
            scanf("%d %d %d", &edge[i].u, &edge[i].v, &edge[i].cost);
            edge[i+1].u=edge[i].v;
            edge[i+1].v=edge[i].u;
            edge[i+1].cost=edge[i].cost;
        }
        for(int i =2*edgenum+1; i <= 2*edgenum+w; i++)//加上虫洞，单向边，负权 
        {
            scanf("%d %d %d", &edge[i].u, &edge[i].v, &edge[i].cost);
            edge[i].cost=-edge[i].cost;
        }
        if(Bellman_Ford())//没有负环 
            printf("NO
");
        else
            printf("YES
");
    }
    return 0;
}