2021牛客暑期多校训练营1 - D - Determine the Photo Position - 题解

Posted 2021-08-17 Tisfy

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2021牛客暑期多校训练营1 - D - Determine the Photo Position

传送门
Time Limit: 1 second
Memory Limit: 262144K

题目描述

You have taken the graduation picture of graduates. The picture could be regarded as a matrix $A$ of $\\times n$ , each element in A is 0 or 1, representing a blank background or a student, respectively.

However, teachers are too busy to take photos with students and only took a group photo themselves. The photo could be regarded as a matrix $B$ of $\\times m$ where each element is 2 representing a teacher.

As a master of photoshop, your job is to put photo B into photo A with the following constraints:
you are not allowed to split, rotate or scale the picture, but only translation.
each element in matrix B should overlap with an element in A completely, and each teacher should overlap with a blank background, not shelter from a student.

Please calculate the possible ways that you can put photo B into photo A.

输入描述:

The first line contains two integers $\\le n,m \\le 2000)$ indicating the size of photos A and B.

In the next $n$ lines,each line contains ${n}$ characters of ‘0’ or ‘1’,representing the matrix $A$ .

The last line contains ${m}$ characters of ‘2’, representing matrix $B$ .

输出描述:

Output one integer in a line, indicating the answer.

样例

样例一

输入

输出

样例二

输入

输出

样例三

输入

输出

题目大意

拍毕业照有 $n\\times n$ 个位置，其中0是空位1是学生
有 $m$ 个老师（一排挨着）
问有多少种方式能把老师正好插入到学生的空位中（老师仍然一排挨着）

解题思路

那就对于学生的每一排，看这一排中有多少个连续的空位。
连续空位 $t$ 大于等于 $m$ 个就有 $t - m + 1$ 种插入方式。
连续空位数小于 $m$ 则不能插入。

比如长度为3的漂亮数组的最大值是 $1 + 3 + 5 = 9$ ，那么8可表示成 $1 + 3 + 4$ 是长度为3的漂亮数组。

AC代码

#include <bits/stdc++.h>
using namespace std;
#define mem(a) memset(a, 0, sizeof(a))
#define dbg(x) cout << #x << " = " << x << endl
#define fi(i, l, r) for (int i = l; i < r; i++)
#define cd(a) scanf("%d", &a)
typedef long long ll;
int getOne(int lx, int m) // 参数：连续的空位数、要插入的老师数  |  返回：有多少种插入方式
{
    if(lx<m)return 0; // 空位数小于老师数，不能插入，插入方式是0
    return lx-m+1; // 空位数≥老师数，插入方式是空位数-老师数+1
}
int main()
{
    int n,m;
    cin>>n>>m;
    int ans=0;
    for(int i=0;i<n;i++)
    {
        string s;
        cin>>s;
        s = '1'+s+'1'; // 前后各添加一个学生，对空位不产生影响，相当于是哨兵，最后必定会遍历到1，遍历到1就把连续的空位累加到结果中。
        int lx=0; // 连续的空位数
        for(int j=1;j<=n+1;j++) // 遍历这一排
        {
            if(s[j]=='0')lx++; // 是空位，连续的空位数就+1
            else // 不是空位，前面的空位（如果有的话）将不连续
            {
                ans+=getOne(lx,m); // 累加到结果中
                lx=0; // 连续的空位数变成0
            }
        }
    }
    cout<<ans<<endl; // 输出结果即可。
    return 0;
}