zoukankan      html  css  js  c++  java
  • hdu 1102 Constructing Roads (kruskal)

    Constructing Roads
    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 28154    Accepted Submission(s): 10717

    Problem Description
    There are N villages, which are numbered from 1 to N, and you should build some roads such that every two villages can connect to each other. We say two village A and B are connected, if and only if there is a road between A and B, or there exists a village C such that there is a road between A and C, and C and B are connected.
    We know that there are already some roads between some villages and your job is the build some roads such that all the villages are connect and the length of all the roads built is minimum.
     
    Input
    The first line is an integer N (3 <= N <= 100), which is the number of villages. Then come N lines, the i-th of which contains N integers, and the j-th of these N integers is the distance (the distance should be an integer within [1, 1000]) between village i and village j.
    Then there is an integer Q (0 <= Q <= N * (N + 1) / 2). Then come Q lines, each line contains two integers a and b (1 <= a < b <= N), which means the road between village a and village b has been built.
     
    Output
    You should output a line contains an integer, which is the length of all the roads to be built such that all the villages are connected, and this value is minimum.
     
    Sample Input
    3
    0 990 692
    990 0 179
    692 179 0
    1
    1 2
     
    Sample Output
    179
     

    C/C++:

     1 #include <cstdio>
     2 #include <climits>
     3 #include <algorithm>
     4 using namespace std;
     5 
     6 int n, m, my_map[110][110], my_pre[110];
     7 struct node
     8 {
     9     int a, b, a_b_dis;
    10 }my_dis[6000];
    11 
    12 bool cmp(node a, node b)
    13 {
    14     return a.a_b_dis < b.a_b_dis;
    15 }
    16 
    17 int my_find(int x)
    18 {
    19     int n = x;
    20     while (n != my_pre[n])
    21         n = my_pre[n];
    22     int i = x, j;
    23     while (n != my_pre[i])
    24     {
    25         j = my_pre[i];
    26         my_pre[i] = n;
    27         i = j;
    28     }
    29     return n;
    30 }
    31 
    32 int my_kruskal()
    33 {
    34     int my_cnt = 0, my_ans = 0;
    35     for (int i = 1; i < n; ++ i)
    36     {
    37         for (int j = i + 1; j <= n; ++ j)
    38         {
    39             my_dis[my_cnt].a = i;
    40             my_dis[my_cnt].b = j;
    41             my_dis[my_cnt].a_b_dis = my_map[i][j];
    42             my_cnt ++;
    43         }
    44     }
    45     sort(my_dis, my_dis + my_cnt, cmp);
    46     for (int i = 0; i < my_cnt; ++ i)
    47     {
    48         int n1 = my_find(my_dis[i].a), n2 = my_find(my_dis[i].b);
    49         if (n1 != n2)
    50         {
    51             my_pre[n1] = n2;
    52             my_ans += my_dis[i].a_b_dis;
    53         }
    54     }
    55     return my_ans;
    56 }
    57 
    58 int main()
    59 {
    60     /**
    61         Date Input Initialize
    62     */
    63     while (~scanf("%d", &n))
    64     {
    65         for (int i = 1; i <= n; ++ i)
    66             for (int j = 1; j <= n; ++ j)
    67                 scanf("%d", &my_map[i][j]);
    68         scanf("%d", &m);
    69         for (int i = 1; i <= n; ++ i)
    70             my_pre[i] = i;
    71         for (int i = 0; i < m; ++ i)
    72         {
    73             int a, b, a_temp, b_temp;
    74             scanf("%d%d", &a, &b);
    75             a_temp = my_find(a), b_temp = my_find(b);
    76             my_pre[a_temp] = b_temp;
    77         }
    78         printf("%d
    ", my_kruskal());
    79     }
    80     return 0;
    81 }
  • 相关阅读:
    Pymsql
    MySQL基础操/下
    MySQL基础操作
    前端学习之jquery/下
    前端学习之jquery
    Python之异常处理
    Python之模块和包导入
    Python之模块
    Python之面向对象上下文管理协议
    Python之面向对象slots与迭代器协议
  • 原文地址:https://www.cnblogs.com/GetcharZp/p/9427525.html
Copyright © 2011-2022 走看看