poj 3259

Wormholes

Time Limit: 2000MS Memory Limit: 65536KTotal 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

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

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.csdn.net/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 0x3f3f3f3fusing 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");    }}