how many tables

本文介绍如何使用Java编程语言和并查集数据结构来解决HDOJ在线评测平台上的问题1213。通过详细解释问题背景、并查集的基本概念和实现步骤,读者将深入了解并查集在实际编程问题中的应用,并掌握解决类似问题的方法。


//java

//http://acm.hdu.edu.cn/showproblem.php?pid=1213
import java.util.Arrays;
import java.util.Scanner;
public class Main {
static Scanner inScanner=new Scanner (System.in);
static int peo;
static int[] rel=new int[1010];
static int count;
static int findrel(int t){
if(rel[t]==-1)
return t;
return findrel(rel[t]);
}
static void addp(int x,int y){
int po1=findrel(x);
int po2=findrel(y);
if(po1!=po2)
rel[po1]=po2;
}
public static void main(String[] args) {
// TODO Auto-generated method stub
while(inScanner.hasNext()){
int num=inScanner.nextInt();
while(num--!=0){
count=0;
peo=inScanner.nextInt();
Arrays.fill(rel, -1);
int fri=inScanner.nextInt();
while(fri--!=0){
int x=inScanner.nextInt();
int y=inScanner.nextInt();
addp(x, y);
}
for(int i=1;i<=peo;i++){
if(rel[i]==-1){
count++;
}
}
System.out.println(count);
}
}
}


}
6-4 朋友聚会 分数 10 作者 杜祥军 单位 青岛大学 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. 函数接口定义: int find(int x); 其中 N 和 D 都是用户传入的参数。 N 的值不超过int的范围; D 是[0, 9]区间内的个位数。函数须返回 N 中 D 出现的次数。 裁判测试程序样例: #include <stdio.h> int pre[1010]; int find(int x); int main() { int t; scanf("%d",&t); while(t--){ int n ,m; scanf("%d%d",&n,&m); for(int i=1;i<=n;i++) pre[i]=i; for(int i=0;i<m;i++){ int x,y; scanf("%d%d",&x,&y); int fx=find(x); int fy=find(y); if(fx!=fy) pre[fx]=fy; } int cnt=0; for(int i=1;i<=n;i++) if(pre[i]==i) cnt++; printf("%d\n",cnt); } return 0; } /* 请在这里填写答案 */ 输入样例: 2 5 3 1 2 2 3 4 5 5 1 2 5 输出样例: 2 4 代码长度限制 16 KB 时间限制 1000 ms 内存限制 32 MB C++ (g++) 1 给出参考代码
10-28
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