Sereja and Bottles-水题有点坑爹

Time Limit: 2000MS

Memory Limit: 262144KB

64bit IO Format: %I64d & %I64u

Description

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 Input

Input

Output

Input

Output

理解题意就可以

/*

Author: 2486

Memory: 812 KB		Time: 60 MS

Language: GNU G++ 4.9.2		Result: Accepted

Public:		No Yes

*/

#include <cstdio>

#include <cstring>

#include <algorithm>

using namespace std;

const int maxn=1000+5;

bool vd[maxn][maxn];

struct fs{

    int a,b;

}fss[maxn];

int main() {

    int b,n;

    scanf("%d",&n);

    memset(vd,false,sizeof(vd));

    for(int i=1; i<=n; i++) {

        scanf("%d%d",&fss[i].a,&fss[i].b);

        vd[fss[i].b][fss[i].a]=true;

    }

    bool flag=true;

    int cnt=0;

    for(int i=1; i<=n; i++) {

        flag=false;

        for(int j=1; j<=n; j++) {

            if(i==j)continue;

            if(fss[i].a==fss[j].b) {

                flag=true;

                break;

            }

        }

        if(flag)cnt++;

    }

    printf("%d\n",n-cnt);

    return 0;

}

巴特西

Sereja and Bottles-水题有点坑爹

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