How Many Tables
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 24265 Accepted Submission(s): 12128
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.
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
Author
Ignatius.L
Source
题意:
就是简单并查集问题的模板。不多说了。
加一句:
英语水水的我,看到这两句就懵了......
代码:
#include<stdio.h>
#define MYDD 1103
int pre[MYDD];
void init(int x) {
for(int j=1; j<=x; j++)
pre[j]=j;
}
int find(int x) { //递归查找根节点
return x==pre[x]? x:find(pre[x]);
}
void combine(int x,int y) {
int fx=find(x);
int fy=find(y);
if(fx!=fy)
pre[fx]=fy;
}
int main() {
int t;
scanf("%d",&t);
while(t--) {
int n,m;
scanf("%d%d",&n,&m);
init(n);
int a,b;
while(m--) {
scanf("%d%d",&a,&b);
combine(a,b);
}
int ans=0;
for(int j=1; j<=n; j++) {
if(pre[j]==j)
ans++;//查找根接点个数
}
printf("%d\n",ans);
}
return 0;
}
并查集是基础。
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本文通过一个具体的并查集问题实例,介绍了并查集的基本概念、实现方法及其应用。通过对问题的逐步分析,展示了如何使用并查集解决实际问题。

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