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
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.
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[N],t[N],r[N];
int st[N];
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;
}