zoukankan      html  css  js  c++  java
  • 九度OJ 1445:How Many Tables

    题目描述:

    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.

    输入:

    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.

    输出:

    For each test case, just output how many tables Ignatius needs at least. Do NOT print any blanks.

    样例输入:
    2
    5 3
    1 2
    2 3
    4 5
    
    5 1
    2 5
    样例输出:
    2
    4
    #include <stdio.h>
    #define N 1002
    int Tree[N];
    int findRoot(int x){
        int ret = x;
        while(Tree[ret] != -1){
            ret = Tree[ret];
        }
        while(Tree[x] != -1){
            int t = Tree[x];
            Tree[x] = ret;
            x = t;
        }
        return ret;
    }
    int main(){
        int t;
        scanf("%d",&t);
        while(t--){
            int n,m;
            scanf("%d%d",&n,&m);
            int i;
            for(i = 1;i <= n;i++)
                Tree[i] = -1;
            for(i = 0;i < m;i++){
                int x,y;
                scanf("%d%d",&x,&y);
                int a = findRoot(x);
                int b = findRoot(y);
                if(a != b){
                    Tree[a] = b;
                }
            }
            int cnt = 0;
            for(i = 1;i <= n;i++){
                if(Tree[i] == -1)cnt++;
            }
            printf("%d
    ",cnt);
        }
    }
  • 相关阅读:
    学习TextKit框架(上)
    UITextView -- 基础备忘
    Quartz2D 备忘 + 学习
    CALayer -- 备忘
    NSURLSession -- 实际开发中运用
    NSURLSession -- 备忘
    Collection View 自定义布局(custom flow layout)
    CSS中一个冒号和两个冒号之间区别
    Chrome插件LiveStyle结合Sublime Text编辑器实现高效可视化开发
    Taking Advantage of HTML5 and CSS3 with Modernizr
  • 原文地址:https://www.cnblogs.com/starryxsky/p/7095568.html
Copyright © 2011-2022 走看看