zoukankan      html  css  js  c++  java
  • How Many Tables(POJ 1213 求连通分量)

    How Many Tables

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 20974    Accepted Submission(s): 10382


    Problem Description
    Today is Ignatius' birthday. He invites a lot of friends. Now it's dinner time. Ignatius wants to know how many tables he needs at least. You have to notice that not all the friends know each other, and all the friends do not want to stay with strangers.

    One important rule for this problem is that if I tell you A knows B, and B knows C, that means A, B, C know each other, so they can stay in one table.

    For example: If I tell you A knows B, B knows C, and D knows E, so A, B, C can stay in one table, and D, E have to stay in the other one. So Ignatius needs 2 tables at least.
     
    Input
    The input starts with an integer T(1<=T<=25) which indicate the number of test cases. Then T test cases follow. Each test case starts with two integers N and M(1<=N,M<=1000). N indicates the number of friends, the friends are marked from 1 to N. Then M lines follow. Each line consists of two integers A and B(A!=B), that means friend A and friend B know each other. There will be a blank line between two cases.
     
    Output
    For each test case, just output how many tables Ignatius needs at least. Do NOT print any blanks.
     
    Sample Input
    2
    5 3
    1 2
    2 3
    4 5
    5 1
    2 5
     
    Sample Output
    2
    4
     
     1 #include <iostream>
     2 #include <cstdio>
     3 #include <cstring>
     4 #include <algorithm>
     5 #include <set>
     6 #include <cmath>
     7 using namespace std;
     8 #define Max 1000+10
     9 int per[Max];
    10 int n,m;
    11 void init()
    12 {
    13     for(int i=1;i<=n;i++)
    14         per[i]=i;
    15 }
    16 int find(int a)
    17 {
    18     if(a==per[a])
    19         return a;
    20     /*int root=a,x=a,j;
    21     while(root!=per[root])    root=per[root];
    22     while(x!=per[x])
    23     {
    24         j=per[x];
    25         per[x]=root;
    26         x=j;
    27     }*/
    28     return per[a]=find(per[a]);
    29 }
    30 int unite(int a,int b)
    31 {
    32     a=find(a);
    33     b=find(b);
    34     if(a!=b)
    35         per[a]=b;
    36     return 0;
    37 }
    38 int main()
    39 {
    40     int t,a,b;
    41     freopen("in.txt","r",stdin);
    42     scanf("%d",&t);
    43     while(t--)
    44     {
    45         int ans=0;
    46         scanf("%d%d",&n,&m);
    47         init();
    48         for(int i=1;i<=m;i++)
    49         {
    50             scanf("%d%d",&a,&b);
    51             unite(a,b);
    52         }
    53         for(int i=1;i<=n;i++)
    54             if(per[i]==i)
    55                 ans++;
    56         printf("%d
    ",ans);
    57     }
    58 }
    Author
    Ignatius.L
  • 相关阅读:
    获取汉字信息(结合正则就可以得到想要的详细啦)
    压缩图片(递归结合pillow)通过改变图片尺寸实现;tinify 需要付费
    实现两个视频同时播放,利用到opencv模块 (线程进程开启)
    切换pip下载源头
    516. 最长回文子序列
    87.扰乱字符串
    Maximum Likelihood ML
    数组右边第一个比当前元素大的数
    4. 寻找两个正序数组的中位数
    min-hash
  • 原文地址:https://www.cnblogs.com/a1225234/p/5181263.html
Copyright © 2011-2022 走看看