poj3259

最新推荐文章于 2022-07-27 13:14:52 发布

翻译最新推荐文章于 2022-07-27 13:14:52 发布 · 374 阅读

文章标签：

#each #numbers #output #path #integer #include

图论同时被 2 个专栏收录

2 篇文章

订阅专栏

spfa算法

1 篇文章

订阅专栏

在FJ的农场探索中，发现了一系列神秘的虫洞。FJ希望利用这些虫洞，从某些地点出发，通过路径和虫洞，返回到出发前的时间点。本文详细介绍了如何使用SPFA算法解决这个问题，以及在特定情况下是否可能实现时间旅行。

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

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

spfa算法中，当出现负环的时候某个节点进出队列的次数一定大于或等于节点数。

#include<iostream>
#include<queue>
#include<algorithm>
using namespace std;
int map[5500][5500];
int dist[5500];
int cnt[5500];
int inqueue[5500];

int spfa(int n) {
    queue<int> Q;
    while (!Q.empty())
        Q.pop();
    memset(inqueue, 0, sizeof (inqueue));
    memset(cnt, 0, sizeof (cnt));
    for (int i = 0; i < 5500; i++)
        dist[i] = 100000000;
    dist[1] = 0;
    inqueue[1] = 1;
    cnt[1]++;
    Q.push(1);
    while (!Q.empty()) {
        int tmp = Q.front();
        Q.pop();
        inqueue[tmp] = 0;
        for (int i = 1; i <= n; i++) {
            if (dist[tmp] + map[tmp][i] < dist[i]) {
                dist[i] = dist[tmp] + map[tmp][i];
                if (!inqueue[i]) {
                    cnt[i]++;
                    if (cnt[i] >= n)
                        return 1;
                    inqueue[i] = 1;
                    Q.push(i);
                }
            }
        }
    }
    return 0;
}

int main() {
    int test_cases;
    cin >> test_cases;
    while (test_cases--) {
        int n, m, w;
        cin >> n >> m >> w;
        for (int i = 0; i <= n; i++)
            for (int j = 0; j <= n; j++)
                map[i][j] = 100000000;
        for (int i = 0; i < m; i++) {
            int s, e, t;
            cin >> s >> e >> t;
            if (map[s][e] > t)
                map[s][e] = map[e][s] = t;
        }
        for (int i = 0; i < w; i++) {
            int s, e, t;
            cin >> s >> e >> t;
            map[s][e] = -t;
        }
        if (spfa(n))
            cout << "YES\n";
        else
            cout << "NO\n";
    }
    return 0;
}