HDOJ1856(More is better)

More is better

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 327680/102400 K (Java/Others)
Total Submission(s): 661    Accepted Submission(s): 247

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.

 

一开始用标准并查集做TLE了n次，就是以下代码:

 

//Time Limit Exceeded 1856 1000MS 78576K 1492 B C++ Xredman 
//Time Limit Exceeded 1856 1000MS 78576K 1790 B C++ Xredman 
#include <iostream>
#include <map>
using namespace std;

const int N =  10000002;

typedef struct
{
    int parent;
    int height;
}Node;

Node boy[N];
int cnt[N];
int n, maxb, minb;

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

int find(int x)
{
    while(boy[x].parent != x)
        x = boy[x].parent;
    return x;
}

void merge(int a, int b)
{
    a = find(a);
    b = find(b);
    if(a == b)
        return ;
    if(boy[a].height == boy[b].height)
    {
        boy[b].parent = a;
        boy[a].height++;
    }
    else if(boy[a].height < boy[b].height)
    {
        boy[b].parent = a;
    }
    else
    {
        boy[a].parent = b;
    }
}

int main()
{
    int a, b, result, i;
    bool flag ;
    while(cin>>n)
    {
        maxb = -1;
        init();
        while(n--)
        {
            scanf("%d%d", &a, &b);
            if(minb == -1 || minb > a) minb = a;
            if(maxb < b) maxb = b;
            merge(a,b);
        }
        for(i = minb; i <= maxb; i++)
        {
            cnt[i] = 0;
        }
        result = 0;
        for(i = minb; i <= maxb; i++)
        {
            a = find(i);
            cnt[a]++;
            if(result < cnt[a])
                result = cnt[a];
        }
        
        cout<<result<<endl;
    }
    return 0;
}

 

后来寻找到一种在“查”时同时“并”时的算法，总算过了！

 

//Accepted    1856    250MS    78592K    990 B    C++    Xredman
#include <iostream>
using namespace std;

const int N =  10000002;

typedef struct
{
    int parent;
    int cnt;
}Node;

Node 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;
        boy[x].parent = y;
        x = tmp;
    }
    return y;
}

void merge(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)
                merge(a,b);
        }
        cout<<result<<endl;
    }
    return 0;
}