More is better
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 327680/102400 K (Java/Others)
Total Submission(s): 18707 Accepted Submission(s):
6873
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.
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)
equals to the maximum number of boys Mr Wang may keep.
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.
#include <iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
#define maxn 100000
using namespace std;
int ran[maxn],per[maxn],via[maxn],num[maxn],n,sum;
void init()
{
int i;
for(i=;i<=maxn;i++)
{
per[i]=i;
num[i] = ;//记录每个根节点所包含的子节点
}
}
int find(int x)//查找函数
{
int t=x;
while(t!=per[t])
t=per[t];
int i=x,j;
while(i!=t)
{
j=per[i];
per[i]=t;
i=j;//压缩路径
}
return t;
}
void join(int x,int y)//合并函数，合并图
{
int fx=find(x);
int fy=find(y);
if(fx!=fy)
{
per[fx] = fy;
num[fy] += num[fx];
sum = max(sum, num[fy]);//求出各个树中子节点最多的
}
else
return ;
}
int main ()
{
int a,b,i,w,n;
while(scanf("%d",&n)!=EOF)
{
if(n == ){
printf("1\n");//特殊坑爹数据
continue;
}
init();
w=n;
sum = ;
while(w--)
{
scanf("%d%d",&a,&b);
via[a]=;
via[b]=;
join(a,b);
}
printf("%d\n", sum);
}
return ;
}