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Wormholes
题目链接：http://poj.org/problem?id=3259
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 backT 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 8Sample Output
NO
YESHint
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.
题意：虫洞是很奇特的，因为它是一个**单向**通道，可让你进入虫洞的前达到目的地！他的N（1≤N≤500）个农场被编号为1..N，之间有M（1≤M≤2500）条路径，W（1≤W≤200）个虫洞。FJ作为一个狂热的时间旅行的爱好者，他要做到以下几点：开始在一个区域，通过一些路径和虫洞旅行，他要回到最开时出发的那个区域出发前的时间。也许他就能遇到自己了:)。为了帮助FJ找出这是否是可以或不可以，他会为你提供F个农场的完整的图（1≤F≤5）。所有的路径所花时间都不大于10000秒，所有的虫洞都回到不大于10000秒之前。
思路：Bellman-Ford算法模板题,判断是否存在负权回路，注意回溯的时间输入的是正的，要转化为负数。
AC代码：
204ms
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<string>
#include<queue>
#include<vector>
#define ll long long
const int INF = 0x3f3f3f3f;
const int MAXN = 1000000 + 5;
using namespace std;
int n, m, w;
double y;
double dist[MAXN];
//double map[110][110];
struct edge{
int st,end;
int cost;
}e[3000];
//int visit[110];
//void dijkstra(int x);
bool Bellman_Ford(int cnt);
int main()
{
int t;
scanf("%d", &t);
while (t--) {
scanf("%d%d%d", &n, &m, &w);
int k = 0;
for (int i = 1; i <= m; ++i) {
int a, b, c;
k++;
scanf("%d%d%d", &a, &b, &c);
e[k].st = a;
e[k].end = b;
e[k].cost = c;
k++;
e[k].st = b;
e[k].end = a;
e[k].cost = c;
}
for (int i = 1; i <= w; ++i) {
k++;
scanf("%d%d%d", &e[k].st, &e[k].end, &e[k].cost);
e[k].cost = -e[k].cost;
}
if (!Bellman_Ford(k))
printf("YES\n");
else
printf("NO\n");
}
}
bool Bellman_Ford(int cnt) {
for (int i = 1; i <= n; ++i) {
dist[i] = INF;
}
dist[1] = 0;
for (int j = 1; j < n; ++j) {
for (int i = 1; i <= cnt; ++i) {
int x = e[i].st, y = e[i].end;
if (dist[y] > dist[x] + e[i].cost) {
dist[y] = dist[x] + e[i].cost;
}
}
}
for (int i = 1; i <= cnt; ++i) {
int x = e[i].st, y = e[i].end;
if (dist[y] > dist[x] + e[i].cost)
return false;
}
return true;
}