POJ 3259 Wormholes
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Wormholes
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. Source
USACO 2006 December Gold
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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;
}