九度oj 1444
2015-08-01 21:46
232 查看
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.
输入:
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)
输出:
The output in one line contains exactly one integer equals to the maximum number of boys Mr Wang may keep.
样例输入:
样例输出:
#include<iostream>
#include<stdio.h>
using namespace std;
#define N 10000001
int tree
;
int findroot(int x)
{
if(tree[x]==-1)
return x;
else
return findroot(tree[x]);
}
int sum
;
int main()
{
int n;
while(cin>>n)
{
memset(tree,-1,sizeof(tree));
for(int i=1;i<N;i++)
{
tree[i]==-1;
sum[i]=1;
}
while(n--!=0)
{
int a,b;
cin>>a>>b;
a=findroot(a);
b=findroot(b);
if(a!=b)
{
tree[a]=b;
sum[b]+=sum[a];
}
}
int ans=1;
for(int i=1;i<=N;i++)
{
if(tree[i]==-1&&sum[i]>ans)
{
ans=sum[i];}
}
cout<<ans<<endl;
}
}
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.
输入:
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)
输出:
The output in one line contains exactly one integer equals to the maximum number of boys Mr Wang may keep.
样例输入:
4 1 2 3 4 5 6 1 6 4 1 2 3 4 5 6 7 8
样例输出:
42
#include<iostream>
#include<stdio.h>
using namespace std;
#define N 10000001
int tree
;
int findroot(int x)
{
if(tree[x]==-1)
return x;
else
return findroot(tree[x]);
}
int sum
;
int main()
{
int n;
while(cin>>n)
{
memset(tree,-1,sizeof(tree));
for(int i=1;i<N;i++)
{
tree[i]==-1;
sum[i]=1;
}
while(n--!=0)
{
int a,b;
cin>>a>>b;
a=findroot(a);
b=findroot(b);
if(a!=b)
{
tree[a]=b;
sum[b]+=sum[a];
}
}
int ans=1;
for(int i=1;i<=N;i++)
{
if(tree[i]==-1&&sum[i]>ans)
{
ans=sum[i];}
}
cout<<ans<<endl;
}
}
相关文章推荐
- 把线性方程组变系数矩阵和常量形式
- 黑马程序员------网络编程(TCP&UDP)
- 公交车比喻
- 说反话(c++实现)
- nutch2.2.1 mysql 建表语句
- 说反话(c++实现)
- Decode Ways
- PHP实例开发(2)PHP通过mail()或Socket发邮件
- EFCode First 导航属性
- 斐波那契数列问题
- pandas groupby (TimeGrouper)重写Q3
- java 微信开发
- 探索angular源码--启动(1)
- EFCode First 导航属性
- 开发 IDE 从 Eclipse 转移到 IntelliJ Idea 中
- 字符串相关
- 使用缓存Memcache存储更新微信access token
- mysql 数据库分区
- Scala入门到精通——第十五节 Case Class与模式匹配(二)
- EFCode First 导航属性