More is better
https://vjudge.net/problem/11137/origin
题目描述
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.
AC代码
#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;
int pd[100010];
int Find(int x)
{
if(pd[x]<0) return x;
else return pd[x] = Find(pd[x]);
}
void Union(int x,int y)
{
int u=Find(x);
int v=Find(y);
if(u!=v)
{
if(pd[u]<pd[v])
{
pd[u]+=pd[v];
pd[v]=u;
}
else
{
pd[v]+=pd[u];
pd[u]=v;
}
}
}
int main()
{
int n,x,y;
while(~scanf("%d",&n))
{
memset(pd,-1,sizeof(pd));
for(int i=1;i<=n;i++)
{
scanf("%d%d",&x,&y);
Union(x,y);
}
int mi=0x7fffffff;
for(int i=0;i<=100000;i++)
{
mi=min(pd[i],mi);
}
cout<<-mi<<endl;
}
}
这道题目直接套用模板会超时，所以在初始化数组的时候不妨把所有的数组初始化为-1，然后联通一个加一个上去，输出时找数组中最小的数就是了