【算法笔记第9.6节-并查集】问题 D: More is better

题目描述

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.

输入

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)

输出

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

样例输入

3
1 3
1 5
2 5
4
3 2
3 4
1 6
2 6

样例输出

4
5

题目说的是现在有一个项目需要一个队伍完成，而且队伍人数越多越好。

现在有10000000个人，给你一些两人之间的联系，让你输出队伍人数最多的人的数量。

#include<stdio.h>
#include<string.h>
int f[10000001];
int num[10000001];
int findFather(int x)
{
    if(x==f[x]) return x;
    else
    {
        int F = findFather(f[x]);
        f[x] = F;
        return F;
    }
}
void Union(int a, int b)
{
    int fa = findFather(a);
    int fb = findFather(b);
    if(fa!=fb)
        f[fa] = fb;
}
int main()
{
    int m;
    while(scanf("%d",&m)!=EOF)
    {
        memset(num, 0, sizeof(num));
        for(int i=1; i<=10000000; i++)
            f[i] = i;
        int a, b;
        for(int i=0; i<m; i++)
        {
            scanf("%d%d",&a, &b);
            Union(a, b);
        }
        int ans = 0;
        for(int i=1; i<=10000000; i++)
        {
            num[findFather(i)]++;
        }
        int maxn = 1;
        for(int i=1; i<=10000000; i++)
        {
            if(maxn < num[i])
                maxn = num[i];
        }
        printf("%d\n", maxn);
    }
    return 0;
}