We have N boxes, numbered 1 through N. At first, box 1 contains one red ball, and each of the other boxes contains one white ball.
Snuke will perform the following M operations, one by one. In the i-th operation, he randomly picks one ball from box xi, then he puts it into box yi.
Find the number of boxes that may contain the red ball after all operations are performed.
Constraints
2≤N≤105
1≤M≤105
1≤xi,yi≤N
xi≠yi
Just before the i-th operation is performed, box xi contains at least 1 ball.
Print the number of boxes that may contain the red ball after all operations are performed.
Just after the first operation, box 1 is empty, box 2 contains one red ball and one white ball, and box 3 contains one white ball.
Now, consider the second operation. If Snuke picks the red ball from box 2, the red ball will go into box 3. If he picks the white ball instead, the red ball will stay in box 2. Thus, the number of boxes that may
contain the red ball after all operations, is 2.
//模拟取球过程 分类讨论 模拟法
#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
struct box
{
int num;
int flag;
}A[100005];
int main()
{
int n,m;
cin>>n>>m;
for(int i=1;i<=n;i++)
{
if(i==1)
{
A[i].flag = 1;
A[i].num = 1;
}
else
{
A[i].flag = 0;
A[i].num = 1;
}
}
int x,y;
for(int i=0;i<m;i++)
{
cin>>x>>y;
if(A[x].flag==1)
{
if(A[x].num==1)
{
A[y].flag = 1;
A[y].num++;
A[x].flag = 0;
A[x].num--;
}
else
{
A[x].num--;
A[y].num++;
A[y].flag = 1;
}
}
else
{
A[x].num--;
A[y].num++;
}
}
int ans = 0;
for(int i=1;i<=n;i++)
{
if(A[i].flag==1)
{
ans++;
}
}
cout<<ans;
}
