几天没有AC了..实在手痒啊，上题。

Wormholes

Time Limit: 2000MS	Memory Limit: 65536K
Total Submissions: 41416	Accepted: 15226

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

1. 正权道路是双向的；
2. 负权道路是单向的；
3. 题目的数据范围貌似小了；
4. 初始化...(大神请无视)

一开始WA了几次，看了discuss说要判连通，到了最后发现其实并不用...

O(nke)做法吧貌似，不太会算..

贴上代码。

/*
Source Code

Problem: 3259		User: aclolicon
Memory: 596K		Time: 1766MS
Language: C++		Result: Accepted
Source Code
*/
#include
#include
#define MAXN 40000
#define INF 0x3f3f3f3f
using namespace std;
struct Edge{
	int to;
	int next;
	int w;
}edge[MAXN];
int head[MAXN];
int cnt = 0;
int n, m, w;

void add(int s, int e, int t){
	edge[cnt].w = t;
	edge[cnt].to = e;
	edge[cnt].next = head[s];
	head[s] = cnt++;
}
bool spfa(int s){
	queue  q;
	q.push(s);
	int cc[MAXN], dis[MAXN];
	bool vis[MAXN];
	for (int i = 0; i < n + 1; i++){
		cc[i] = 0;
		dis[i] = INF;
		vis[i] = 0;
	}
	dis[s] = 0;
	bool flag = 0;
	while(!q.empty()){
		int x = q.front();
		q.pop();
		vis[x] = 0;
		for (int i = head[x]; i != -1 && !flag; i = edge[i].next){
			int to = edge[i].to;
			if (dis[x] + edge[i].w < dis[to]){
				dis[to] = dis[x] + edge[i].w;
				if (!vis[to]){
					vis[to] = 1;
					q.push(to);
					cc[to]++;
					if (cc[to] >= n){
						flag = 1;
						break;
					}
				}
			}
		}
		if (flag) break;
	}
	if (flag) return 0;
	return 1;
}

void init(){
	for (int i = 0; i < n + 1; i++){
		edge[i].to = -1;
		edge[i].w = 0;
		edge[i].next = -1;
		head[i] = -1;
	}
}

void solve(){
	bool flag = 0;
	cnt = 0;
	int s, e, t;
	for (int i = 0; i < m; i++){
		scanf("%d%d%d", &s, &e, &t);
		add(s, e, t);
		add(e, s, t);
	}
	for (int i = 0; i < w; i++){
		scanf("%d%d%d", &s, &e, &t);
		add(s, e, -t);
	}
	for (int i = 1; i <= n; i++){
		if (!spfa(i)) {
			flag = 1;
			break;
		}
	}
	if (flag) printf("YESn");
	else printf("NOn");	
}
 
int main(){
	int f;
	scanf("%d", &f);
	while(f--) {
		scanf("%d%d%d", &n, &m, &w);
		init();
		solve();
	}
	return 0;
}