【杭电】[1856]More is better

More is better

Time Limit: 5000/1000 MS (Java/Others)

Memory Limit: 327680/102400 K (Java/Others)

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.

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要求集合中的元素的最大值 这里可以借助ran数组储存树的高度 (详见并查集防树形退化讲解) http://www.boiltask.com/blog/?p=608 在合并的时候可以对最大高度进行记录

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#include<stdio.h>
int par[10000020],ran[10000020];
int res;
int find(int m) {
    if(m==par[m])
        return m;
    else
        return par[m]=find(par[m]);
}
void unite(int x,int y) {
    x=find(x);
    y=find(y);
    if(x==y)
        return ;
    else {
        if(ran[x]<ran[y]) {
            par[x]=y;
            ran[y]+=ran[x];
            if(ran[y]>res)
                res=ran[y];
        } else {
            par[y]=x;
            ran[x]+=ran[y];
            if(ran[x]>res)
                res=ran[x];
        }
    }
}
int main() {
    int n;
    while(scanf("%d",&n)!=EOF) {
        for(int i=1; i<=10000000; i++) {
            ran[i]=1;
            par[i]=i;
        }
        res=1;
        for(int i=0; i<n; i++) {
            int a,b;
            scanf("%d %d",&a,&b);
            unite(a,b);
        }
        printf("%d
",res);
    }
    return 0;
}

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题目地址:【杭电】[1856]More is better

查看原文：http://www.boiltask.com/blog/?p=1972