Silver Cow Party

最新推荐文章于 2018-04-25 17:03:02 发布

原创最新推荐文章于 2018-04-25 17:03:02 发布 · 271 阅读

0 ·

CC 4.0 BY-SA版权

算法专栏收录该内容

28 篇文章

订阅专栏

本文介绍如何使用Dijkstra算法解决N个农场中每头牛到特定农场参加聚会并返回各自农场所需的最长行走时间问题。通过两次运行Dijkstra算法分别求得前往聚会地点和返回的最短路径。

题目：One cow from each of N farms (1 ≤ N ≤ 1000) conveniently numbered 1..N is going to attend the big cow party to be held at farm #X (1 ≤ X ≤ N). A total of M (1 ≤ M ≤ 100,000) unidirectional (one-way roads connects pairs of farms; road i requires Ti (1 ≤ Ti ≤ 100) units of time to traverse.
Each cow must walk to the party and, when the party is over, return to her farm. Each cow is lazy and thus picks an optimal route with the shortest time. A cow’s return route might be different from her original route to the party since roads are one-way.
Of all the cows, what is the longest amount of time a cow must spend walking to the party and back?
思路本题采用了Dijkstra算法（单源最短路径），首先要清楚最短路径的最优子结构：如果P(i,j)={Vi….Vk..Vs…Vj}是从顶点i到j的最短路径，k和s是这条路径上的一个中间顶点，那么P(k,s)必定是从k到s的最短路径。而Dijkstra算法基于此原理，进行最短路径的递增，并最终生成最短路径。对于原节点，选择与其相邻且路径最短的节点，dist[i] = dist[k] + map[k][i]，后面的节点依据这种思路进行递增。
Dijkstra算法参考博客：http://www.cnblogs.com/dolphin0520/archive/2011/08/26/2155202.html

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
const int MAXN = 1010;
const int INF = 0x3f3f3f3f;
bool visited[MAXN];
void Dijkstra(int map[][MAXN], int dist[], int n, int v0)
{
    for (int i = 1; i <= n; i++)
    {
        dist[i] = INF;
        visited[i] = false;
    }
    dist[v0] = 0;
    for (int j = 0; j<n; j++)
    {
        int k = -1;
        int Min = INF;
        for (int i = 1; i <= n; i++)
            if (!visited[i] && dist[i]<Min)
            {
                Min = dist[i];
                k = i;
            }
        if (k == -1)break;
        visited[k] = true;
        for (int i = 1; i <= n; i++)
            if (!visited[i] && dist[k] + map[k][i]<dist[i])
                dist[i] = dist[k] + map[k][i];
    }
}
int dist1[MAXN];
int dist2[MAXN];
int map[MAXN][MAXN];
int main()
{
    int N, M, X;
    int u, v, w;
    scanf("%d%d%d", &N, &M, &X) == 3; 

    for (int i = 1; i <= N; i++)
        for (int j = 1; j <= N; j++)
        {
            if (i == j)map[i][j] = 0;
            else map[i][j] = INF;
        }
    while (M--)
    {
        scanf("%d%d%d", &u, &v, &w);
        map[u][v] = min(map[u][v], w);
    }
    Dijkstra(map, dist1, N, X);
    for (int i = 1; i <= N; i++)
        for (int j = 1; j<i; j++)
            swap(map[i][j], map[j][i]);
    Dijkstra(map, dist2, N, X);
    int ans = 0;
    for (int i = 1; i <= N; i++)
        ans = max(ans, dist1[i] + dist2[i]);
    printf("%d\n", ans);

    return 0;
}