More is better(并差集)

More is better

Time Limit : 5000/1000ms (Java/Other)   Memory Limit : 327680/102400K (Java/Other)

Total Submission(s) : 1   Accepted Submission(s) : 1

Problem Description

Mr Wang wants some boys to help him with a project. Because the project is rather complex, the more boys come, the better it will be. Of course there are certain requirements.

Mr Wang selected a room big enough to hold the boys. The boy who are not been chosen has to leave the room immediately. There are 10000000 boys in the room numbered from 1 to 10000000 at the very beginning. After Mr Wang's selection any two of them who are still in this room should be friends (direct or indirect), or there is only one boy left. Given all the direct friend-pairs, you should decide the best way.

 

 

Input

The first line of the input contains an integer n (0 ≤ n ≤ 100 000) - the number of direct friend-pairs. The following n lines each contains a pair of numbers A and B separated by a single space that suggests A and B are direct friends. (A ≠ B, 1 ≤ A, B ≤ 10000000)

 

 

Output

The output in one line contains exactly one integer equals to the maximum number of boys Mr Wang may keep.

 

 

Sample Input

4 1 2 3 4 5 6 1 6 4 1 2 3 4 5 6 7 8

 

 

Sample Output

4 2 [hint] A and B are friends(direct or indirect), B and C are friends(direct or indirect), then A and C are also friends(indirect). In the first sample {1,2,5,6} is the result. In the second sample {1,2},{3,4},{5,6},{7,8} are four kinds of answers. [/hint]

王老师要找一些男生帮助他完成一项工程。要求最后挑选出的男生之间都是朋友关系，可以说直接的，也可以是间接地。问最多可以挑选出几个男生（最少挑一个)。

 思路：加一个数组记录数的元素个数；剩下的并差集解决；

代码：

#include<stdio.h>
int relate[10000010],num[10000010];//num[]记录个数；
int min;
int find(int x){
    int r=x;
    while(r!=relate[r])r=relate[r];
    int i=x,j;
    while(i!=r)j=relate[i],relate[i]=r,i=j;
    return r;
}
void initial(){
    for(int i=1;i<=10000000;++i)relate[i]=i,num[i]=1;
}
void merge(int x,int y){
    int f1,f2;
    f1=find(x);f2=find(y);
    if(f1!=f2)relate[f1]=f2,num[f2]+=num[f1];
    min=min>num[f2]?min:num[f2];
}
int main(){
    int n,temp1,temp2;
    while(~scanf("%d",&n)){
        min=1;initial();
        while(n--){
            scanf("%d%d",&temp1,&temp2);
            merge(temp1,temp2);
        }
        printf("%d\n",min);
    }
    return 0;
}