zoukankan      html  css  js  c++  java
  • HDU 2068 Choose the best route

    http://acm.hdu.edu.cn/showproblem.php?pid=2680

    Problem Description
    One day , Kiki wants to visit one of her friends. As she is liable to carsickness , she wants to arrive at her friend’s home as soon as possible . Now give you a map of the city’s traffic route, and the stations which are near Kiki’s home so that she can take. You may suppose Kiki can change the bus at any station. Please find out the least time Kiki needs to spend. To make it easy, if the city have n bus stations ,the stations will been expressed as an integer 1,2,3…n.
     
    Input
    There are several test cases. 
    Each case begins with three integers n, m and s,(n<1000,m<20000,1=<s<=n) n stands for the number of bus stations in this city and m stands for the number of directed ways between bus stations .(Maybe there are several ways between two bus stations .) s stands for the bus station that near Kiki’s friend’s home.
    Then follow m lines ,each line contains three integers p , q , t (0<t<=1000). means from station p to station q there is a way and it will costs t minutes .
    Then a line with an integer w(0<w<n), means the number of stations Kiki can take at the beginning. Then follows w integers stands for these stations.
     
    Output
    The output contains one line for each data set : the least time Kiki needs to spend ,if it’s impossible to find such a route ,just output “-1”.
     
    Sample Input
    5 8 5
    1 2 2
    1 5 3
    1 3 4
    2 4 7
    2 5 6
    2 3 5
    3 5 1
    4 5 1
    2
    2 3
    4 3 4
    1 2 3
    1 3 4
    2 3 2
    1
    1
     
    Sample Output
    1
    -1
     
    时间复杂度:$O(2 * n ^ 2)$
    代码:
    #include <bits/stdc++.h>
    using namespace std;
    
    #define inf 0x3f3f3f3f
    int n, m, s, w;
    int kw;
    int mp[1111][1111];
    int st, en, cost;
    int vis[1111], dis[1111];
    
    void dijkstra(int act) {
        memset(dis, inf, sizeof(dis));
        memset(vis, 0, sizeof(vis));
        
        dis[act] = 0;
        int temp=act;
        for(int i = 1; i < n; i ++) {
            int minn = inf;
            for(int j = 1; j <= n; j ++) {
                if(dis[j] < minn && vis[j] == 0) {
                    minn = dis[j];
                    temp = j;
                }
            }
            vis[temp] = 1;
            for(int k = 1; k <= n; k ++) {
                if(dis[k] > dis[temp] + mp[temp][k])
                    dis[k] = dis[temp] + mp[temp][k];
            }
        }
    }
    
    int main() {
        while(~scanf("%d%d%d", &n, &m, &s)) {
            memset(mp, inf, sizeof(mp));
            for(int i = 1; i <= m; i ++) {
                scanf("%d%d%d", &st, &en, &cost);
                mp[en][st] = min(cost,mp[en][st]);
            }
            scanf("%d", &w);
            dijkstra(s);
            int ans=inf;
            for(int i = 1; i <= w; i ++) {
                scanf("%d", &kw);
                ans = min(dis[kw],ans);
            }
            if(ans < inf)
                printf("%d
    ", ans);
            else
                printf("-1
    ");
        }
        return 0;
    }
    

      

  • 相关阅读:
    C语言 简单的队列(数组队列)
    C语言 复杂的栈(链表栈)
    C语言 简单的栈
    C语言 链表的使用(链表的增删查改,链表逆转,链表排序)
    C语言 常用的时间函数
    C语言 百炼成钢17
    C语言 const常量讲解
    ndroid如何监听开机广播和关机广播
    Android判断GPS是否开启和强制帮用户打开GPS
    android java.lang.IllegalArgumentException: Comparison method violates its general contract! 问题
  • 原文地址:https://www.cnblogs.com/zlrrrr/p/9574352.html
Copyright © 2011-2022 走看看