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CodeForces - 315A| 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 ai, besides, you can use it to open other bottles of brand bi. 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 ai,?bi(1?≤?ai,?bi?≤?1000)— the description of the i-th bottle.
Output
In a single line print a single integer — the answer to the problem.
Sample Input
4 1 1 2 2 3 3 4 4
4
4 1 2 2 3 3 4 4 1
0
理解题意即可
/*
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;
}版权声明：本文为博主原创文章，未经博主允许不得转载。
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原文地址：http://blog.csdn.net/qq_18661257/article/details/47140603
