HDU1865--More is better(统计并查集的秩(元素个数))

More is better

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 327680/102400 K (Java/Others)
Total Submission(s): 25168    Accepted Submission(s): 9045

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.

 

Author

lxlcrystal@TJU

 

Source

HDU 2007 Programming Contest - Final

 

这道题目需要统计并查集的秩

只需要加一个数组记录根节点的秩, 然后在合并时, 顺带着将秩也合并即可

#include <cstdio>
#include <cstdlib>
#include <climits>

const int MAX = 10000005;
int pre[MAX],rank[MAX],maxx;


void init(){
    int i;
    for(i=1;i<MAX;++i){
        pre[i] = i;
        rank[i] = 1;
    }
}

int root(int x){
    if(x!=pre[x]){
        pre[x] = root(pre[x]);
    }
    return pre[x];
}

void merge(int x,int y){
    int fx = root(x);
    int fy = root(y);
    if(fx!=fy){
        pre[fx] = fy;
        rank[fy] += rank[fx];//更改对应元素的秩
        if(rank[fy]>maxx)maxx = rank[fy];//更新最大值
    }
}


int main(){
    //freopen("in.txt","r",stdin);
    int i,n,x,y;
    while(scanf("%d",&n)!=EOF){
        if(n==0){
            printf("1
");
            continue;
        }
        init();
        maxx = INT_MIN;
        for(i=0;i<n;++i){
            scanf("%d %d",&x,&y);
            merge(x,y);
        }
        printf("%d
",maxx);
    }
    return 0;
}