HDU 1856 More is better

More is better

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 327680/102400 K (Java/Others)
Total Submission(s): 5725 Accepted Submission(s): 2127


[align=left]Problem Description[/align]
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.

[align=left]Input[/align]
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)

[align=left]Output[/align]
The output in one line contains exactly one integer equals to the maximum number of boys Mr Wang may keep.

[align=left]Sample Input[/align]

4
1 2
3 4
5 6
1 6
4
1 2
3 4
5 6
7 8

[align=left]Sample Output[/align]

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.

[align=left]Author[/align]
lxlcrystal@TJU

[align=left]Source[/align]
HDU 2007 Programming Contest - Final

[align=left]Recommend[/align]
lcy

Statistic | Submit | Discuss | Note

//因为A,B比较大，所以数据要进行处理，因为n在100000以内，所以开个200000左右的数组，将A,B进行从新编号，然后再进行并查集处理，求最大连通块

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#define N 200003
using namespace std;
struct node
{
int a,b;
};
node te[100003];
int n,m,Max;
int f
,t
,r
;
int st
;
int bf(int x)
{
int l=1,r=m,mid;
while(l<=r)
{
mid=(l+r)>>1;
if(st[mid]>x) r=mid-1;
else if(st[mid]<x) l=mid+1;
else
return mid;
}
}
int find_f(int x)
{
if(x!=f[x])
{
return f[x]=find_f(f[x]);
}
return x;
}
void union_set(int x,int y)
{
x=find_f(x);
y=find_f(y);
if(x==y) return;
if(r[x]>r[y])
{
f[y]=x;
t[x]+=t[y];
if(t[x]>Max) Max=t[x];
}
else if(r[x]<r[y])
{
f[x]=y;
t[y]+=t[x];
if(t[y]>Max) Max=t[y];
}
else
{
f[y]=x;
r[x]++;
t[x]+=t[y];
if(t[x]>Max) Max=t[x];
}
}
int main()
{
int i,j,a,b;
while(scanf("%d",&n)!=EOF)
{
if(n==0) //开始这里没有处理，WA一次
{printf("1\n");continue;}
j=1;Max=0;
for(i=1;i<=n;i++)
{
scanf("%d%d",&te[i].a,&te[i].b);
st[j]=te[i].a;j++;
st[j]=te[i].b;j++;
}
j=2*n;
sort(st+1,st+j+1);//调试时这里写了东西忘记删除、WA一次、悲剧
m=1;
for(i=2;i<=j;i++)
{
if(st[i]!=st[i-1])
st[++m]=st[i];
}
for(i=1;i<=m;i++)
f[i]=i,r[i]=0,t[i]=1;
for(i=1;i<=n;i++)
{
a=bf(te[i].a);
b=bf(te[i].b);
union_set(a,b);
}
printf("%d\n",Max);
}
return 0;
}