杭电1856--More is better

More is better

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 327680/102400 K (Java/Others)
Total Submission(s): 18656    Accepted Submission(s): 6860

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

 

 

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//节点最多的并查集的节点数； (这段代码有一个优化， 根据数节点数不同在合并节点时进行区分)

ac 代码：

#include <cstdio>
#include <iostream>
using namespace std;

const int N = 10000002;

struct node
{
    int parent;
    int cnt; 
} boy[N];

int n;
int result;

void init()
{
    for(int i=0; i<N; i++)
    {
        boy[i].parent = i;
        boy[i].cnt = 1;
    }
}

int find(int x)
{
    int y = x, tmp;
    while(boy[y].parent != y)
        y = boy[y].parent;
    while(x != y)
    {
        tmp = boy[x].parent;     // 暂存父节点； 
        boy[x].parent = y;        // 父节点指向根；
        x = tmp;               // x指向根； 
    }
    return y;
}

void mercy(int a, int b)     //******** ***********//
{
    if(boy[a].cnt > boy[b].cnt)
    {
        boy[b].parent = a;
        boy[a].cnt += boy[b].cnt;
        if(result < boy[a].cnt)
            result = boy[a].cnt;
    }
    else
    {
        boy[a].parent = b;
        boy[b].cnt += boy[a].cnt;
        if(result < boy[b].cnt)
            result = boy[b].cnt;
    }
}


int main()
{
    int a, b, i;
//    bool flag;
    while(cin >> n)
    {
        result = 1;
        init();
        while(n--)
        {
            scanf("%d %d", &a, &b);
            a = find(a);
            b = find(b);
            if(a != b)
            mercy(a, b);
        }
        cout << result << endl; 
    }
    return 0;    
} 

 //又敲了一遍。

#include <cstdio>
#include <iostream>
using namespace std;
const int N = 10000010;
int father[N], have[N], result;
void init()
{
    result = 1;
    for(int i=1; i<N; i++)
    {
        father[i] = i;
        have[i] = 1;
    }
}

int find(int a)
{
    int  r, j, k;
    r = a; 
    while(r != father[r])
        r = father[r];
    //father[a] = a;
    j = a;
    while(j != r)
    {
        k = father[j];
        father[j] = r;
        j = k;
    }
    return r;
}

void mercy(int a,int b)  //非递归压缩路径过， 
{
    int q = find(a);
    int p = find(b);
    if(q != p)
    {
        father[q] = p;
        have[p] += have[q];
        //printf("%d\n", have[p]);
        if(have[p] > result)
            result = have[p];
    }
}

int main()
{
    int a, b, n;
    while(~scanf("%d", &n))
    {
        init();
        while(n--)
        {
            scanf("%d %d", &a, &b);
            mercy(a, b);
        }
        printf("%d\n", result);
    }
    return 0;
}