More is better

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 327680/102400 K (Java/Others)
Total Submission(s): 26832    Accepted Submission(s): 9611


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 [email protected]
Source
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並查集的裸題，題目大意：老師想讓幾個男同學來做一個專案，這些同學必須直接或間接的為朋友關係，求最多招幾個男同學。

思路：這裡集合的數量不確定，目的是求每個集合的元素數量，再加一個num陣列記錄每個集合的元素，初始為1，合併時相加即可。

試了一下不同輸入方式的時間，第一行為cin加取消同步的情況，第二行為cin不加取消同步，第三種是scanf加萬能標頭檔案，第四種是scanf不加萬能標頭檔案。

可以看出第二種直接超時，最優的方案還是寫檔名加scanf。

附上AC程式碼：

#include<bits/stdc++.h>

using namespace std;
const int maxn=10000000+5;
int par[maxn],num[maxn];
int n,f1,f2;
int maxd,mind,maxcnt;

void init()
{
    for(int i=0;i<maxn;i++)
        par[i]=i,num[i]=1;
}

int find(int a)
{
    if(a==par[a]) return a;
    return par[a]=find(par[a]);
}

void unite(int a,int b)
{
    a=find(a);
    b=find(b);
    if(a==b)return ;
    par[a]=b;
    num[b]+=num[a];
}

int main()
{
    while(~scanf("%d",&n))
    {
        if(n==0){printf("1\n");continue;}
        init();
        maxd=INT_MIN;
        mind=INT_MAX;
        maxcnt=0;
        for(int i=0;i<n;i++)
        {
            scanf("%d%d",&f1,&f2);
            maxd=max(maxd,max(f1,f2));
            mind=min(mind,min(f1,f2));
            unite(f1,f2);
        }
        for(int i=mind;i<=maxd;i++)
        {
                maxcnt=max(maxcnt,num[i]);
        }
        printf("%d\n",maxcnt);
    }
    return 0;
}