poj 3259

Wormholes

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 50025		Accepted: 18449

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

思路:可以用弗洛伊德算法解决，这里主要说spfa（队列优化）+ 链式前向星解决此问题，关于链式前向星：

http://blog.youkuaiyun.com/acdreamers/article/details/16902023

 写的非常详细，在给出的顶点中只要出现一个，能够回到原点的，就直接输出结果就行。所以最坏的情况，要调用n次spfa算法。

代码：

#include<cstdio>
#include<cstring>
#include<queue>
#define INF 0x3f3f3f3f
using namespace std;
struct Node {
    int to;
    int weight;
    int next;
};
Node node[5500];
int count = 0;
int n, m, t;
int head[505];
int dis[505];
int visited[505];
int flash[505];
queue<int> q;
void add(int u, int v, int w) {
    node[count].to = v;
    node[count].weight = w;
    node[count].next = head[u];
    head[u] = count++;
}

void spfa(int a) {
    int t;
    memset(visited, 0, sizeof(visited));
    memset(flash, 0, sizeof(flash));
    for(int i = 1; i <= n; i++)
        dis[i] = INF;
    dis[a] = 0;
    while(!q.empty())
        q.pop();
    q.push(a);
    while(!q.empty()) {
        t = q.front();
        q.pop();
        visited[t] = 0;
        for(int k = head[t]; k != -1; k = node[k].next) {
            if(dis[node[k].to] > dis[t] + node[k].weight) {
                dis[node[k].to] = dis[t] + node[k].weight;
                if(!visited[node[k].to]) {
                    q.push(node[k].to);
                    visited[node[k].to] = 1;
                    flash[node[k].to]++;
                    if(flash[node[k].to] > n)   //重要，当优化大于n次后，就说明出现了环， 应该退出。
                        return;
                }
            }
        }
    }
}

int main() {
    int c, i, u, v, w;
    scanf("%d", &c);
    while(c--) {
        count = 0;
        memset(head, -1, sizeof(head));
        scanf("%d%d%d", &n, &m, &t);
        for(i = 1; i <= n; i++)
            dis[i] = INF;
        dis[1] = 0;
        for(i = 0; i < m; i++) {
            scanf("%d%d%d", &u, &v, &w);
            add(u, v, w);
            add(v, u, w);
        }
        for(i = 0; i < t; i++) {
            scanf("%d%d%d", &u, &v, &w);
            add(u, v, -w);
        }
        for(i = 1; i <= n; i++) {
            spfa(i);
            if(dis[i] < 0)
                break;
        }

        if(i != n + 1)
            printf("YES\n");
        else
            printf("NO\n");
    }
}