CF 315A(Sereja and Bottles-开易拉罐)_开易拉罐 self opening soda-优快云博客

Sereja和他的朋友们在野餐时遇到了没有开瓶器的问题，他们需要找到不能打开的瓶子数量。通过使用特定品牌的瓶子来开启其他品牌的瓶子，Sereja希望找出无法开启的瓶子总数。

Sereja and his friends went to a picnic. The guys had n soda bottles just for it. Sereja forgot the bottle opener as usual, so the guys had to come up with another way to open bottles.

Sereja knows that the i-th bottle is from brand a_i, besides, you can use it to open other bottles of brand b_i. You can use one bottle to open multiple other bottles. Sereja can open bottle with opened bottle or closed bottle.

Knowing this, Sereja wants to find out the number of bottles they've got that they won't be able to open in any way. Help him and find this number.

Input

The first line contains integer n (1 ≤ n ≤ 100) — the number of bottles. The next n lines contain the bottles' description. The i-th line contains two integers a_i, b_i (1 ≤ a_i, b_i ≤ 1000) — the description of the i-th bottle.

Output

In a single line print a single integer — the answer to the problem.

Sample test(s)

input
4
1 1
2 2
3 3
4 4

output
4

input
4
1 2
2 3
3 4
4 1

output
0

水题，直接数。注意b[i]对a[j]。

考场上有一大堆人这题各种卖萌，专业Hack30年……

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define MAXN (100+10)
#define MAXM (1000+10)
int a[MAXN],b[MAXN],n;
bool b2[MAXN]={0};
int main()
{
//	freopen(".in","r",stdin);
//	freopen(".out","w",stdout);
    memset(b2,0,sizeof(b2));
    cin>>n;
    For(i,n) cin>>a[i]>>b[i];
    For(i,n)
        For(j,n)
        {
            if (i!=j&&b[i]==a[j]) b2[j]=1;
        }
    int ans=0;
    For(i,n) ans+=b2[i];
    cout<<n-ans<<endl;
	return 0;
}