POJ 3259 Wormholes

Wormholes

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 40421 Accepted: 14805

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

23 3 11 2 21 3 42 3 13 1 33 2 11 2 32 3 43 1 8

Sample Output

NOYES

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

点击打开题目链接POJ3259

题意：N个点，有 M 条双向道路（花费时间为T），W 条单向道路（花费时间为-T），判断是否存在负环

这里我们用 Bellman-Ford 算法先求最短路

如果没有负环，在迭代 N 次后将求出最短路，否则，最短路将不存在（沿着负环一直走，路就会更短）

代码：

#include <cstdio>#include <iostream>using namespace std;const int MAXN = 500 + 10;const int MAXM = 5000 + 1000;const int INF = 1<<30;int t, numOfPath;int N, M, W;int minLength[MAXN];struct road{    int Begin, End, Length;} Road[MAXM];//保存一条从点 u 到点 v 长度为 w 的路径void addPath(int u, int v, int w){    Road[numOfPath].Begin = u;    Road[numOfPath].End = v;    Road[numOfPath++].Length = w;}//Bellman-Ford算法判断是否有负环bool Bellman_Ford(){    for (int i = 1; i <= N; i++) minLength[i] = INF;    minLength[1] = 0;    for (int i = 0; i < N; i++)         //求最短路    {        for (int j = 0; j < numOfPath; j++)        {            int u = Road[j].Begin;            int v = Road[j].End;            int w = Road[j].Length;            if (minLength[u] + w < minLength[v])    //松弛                minLength[v] = minLength[u] + w;        }    }    for (int i = 0; i < N; i++)    {        for (int j = 0; j < numOfPath; j++)        {            int u = Road[j].Begin;            int v = Road[j].End;            int w = Road[j].Length;            if (minLength[u] + w < minLength[v])     //若还能进行松弛，说明不存在最短路而存在负环                return true;        }    }    return false;}int main(){    scanf("%d", &t);    while (t--)    {        scanf("%d%d%d", &N, &M, &W);        numOfPath = 0;        int u, v, w;        for (int i = 0; i < M; i++)     //双向道路        {            scanf("%d%d%d", &u, &v, &w);            addPath(u, v, w);            addPath(v, u, w);        }        for (int i = 0; i < W; i++)     //单向虫洞        {            scanf("%d%d%d", &u, &v, &w);            addPath(u, v, -w);        }        if (Bellman_Ford() == true) printf("YES\n");        else printf("NO\n");    }    return 0;}