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5582: Box and Ball

时间限制: 2 Sec 内存限制: 256 MB
提交: 276 解决: 93
[提交][状态][讨论版][命题人:admin]

题目描述

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.

输入

The input is given from Standard Input in the following format:

N M
x1 y1
:
xM yM

输出

Print the number of boxes that may contain the red ball after all operations are performed.

样例输入

3 2
1 2
2 3

样例输出

提示

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;
}