HDU

More is betterHDU - 1856

  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

题意：给定一些人的认识关系，判断认识人最多的人数是多少。

题解：看题是并查集的做法，加一个数组判断一下有多少认识即可，因为数据太大，用cin会TLE，并且初始化ans应该为1（考虑n==0的情况），或者初始化为1，在最后遍历判断。

AC代码：

#include<iostream>#include<cstdio>#include<cmath>using namespace std;const int maxn = 10000007;int n,ans;int fa[maxn];int sum[maxn];int fi(int x){    return x==fa[x]?x:fa[x]=fi(fa[x]);}void intc(){    for(int i=1;i<=maxn;i++)    {        fa[i]=i;        sum[i]=1;    }}int main(){    while(cin>>n)    {        intc();        ans=0;        for(int i=1;i<=n;i++)        {            int x,y;            cin>>x>>y;            int x1=fi(x);            int y1=fi(y);            if(x1!=y1)            {                fa[x1]=y1;                sum[y1]+=sum[x1];                //ans=max(ans,sum[y1]);//这里如果初始化为0会WA掉。            }        }       for(int i=1;i<=maxn;i++)ans=max(ans,sum[i]);        cout<<ans<<endl;    }    return 0;}