8VC Venture Cup 2016 - Final Round (Div. 2 Edition) A
2 seconds
256 megabytes
standard input
standard output
Paul is at the orchestra. The string section is arranged in an r × c rectangular grid and is filled with violinists with the exception of n violists. Paul really likes violas, so he would like to take a picture including at least k of them. Paul can take a picture of any axis-parallel rectangle in the orchestra. Count the number of possible pictures that Paul can take.
Two pictures are considered to be different if the coordinates of corresponding rectangles are different.
The first line of input contains four space-separated integers r, c, n, k (1 ≤ r, c, n ≤ 10, 1 ≤ k ≤ n) — the number of rows and columns of the string section, the total number of violas, and the minimum number of violas Paul would like in his photograph, respectively.
The next n lines each contain two integers xi and yi (1 ≤ xi ≤ r, 1 ≤ yi ≤ c): the position of the i-th viola. It is guaranteed that no location appears more than once in the input.
Print a single integer — the number of photographs Paul can take which include at least k violas.
2 2 1 1
1 2
4
3 2 3 3
1 1
3 1
2 2
1
3 2 3 2
1 1
3 1
2 2
4
We will use '*' to denote violinists and '#' to denote violists.
In the first sample, the orchestra looks as follows
*#
**
Paul can take a photograph of just the viola, the 1 × 2 column containing the viola, the 2 × 1 row containing the viola, or the entire string section, for 4 pictures total.
In the second sample, the orchestra looks as follows
#*
*#
#*
Paul must take a photograph of the entire section.
In the third sample, the orchestra looks the same as in the second sample.
题意: r*c的矩阵 n个位置特殊 并给出坐标 问能取多少个子矩阵使得其中特殊位置的个数最小为k
题解: 暴力 矩阵最多10*10 子矩阵 数量很小 (水)
#include<iostream>
#include<cstring>
#include<cstdio>
#include<map>
using namespace std;
int r,c,n,k;
int mp[][];
int main()
{
scanf("%d %d %d %d",&r,&c,&n,&k);
for(int i=; i<=n; i++)
scanf("%d %d",&mp[i][],&mp[i][]);
int re=;
for(int i=; i<=r; i++)
{
for(int j=; j<=c; j++)
{
for(int kk=i; kk<=r; kk++)
{
for(int g=j; g<=c; g++)
{
int jishu=;
for(int m=; m<=n; m++)
{
if(mp[m][]<=kk&&mp[m][]>=i&&mp[m][]<=g&&mp[m][]>=j)
jishu++;
}
if(jishu>=k)
re++;
}
}
}
}
printf("%d\n",re);
return ;
}
最新文章
- Qt : QProcess
- NoClassDefFoundError: javax/servlet/jsp/jstl/core/Config
- 10月14日上午PHP环境搭建
- java 字节流和字符流的区别 转载
- C#中跨线程读取控件值、设置控件值
- JDBC中DAO事务函数模版
- 为什么和其他语言相比C语言是快速的语言
- 9.java.lang.ClassCastException
- Java解惑七:很多其它类之谜
- 基于jquery的城市选择插件
- sublime下配置C/C++运行环境
- Android版本分布数据源
- IntelliJ IDEA 自定义方法注解模板
- javascript中startswith和endsWidth 与 es6中的 startswith 和 endsWidth
- Django系列(一)
- python的安装和配置
- Berlekamp-Massey算法学习笔记
- HDU - 5340 Three Palindromes(manacher算法)
- JavaMail发送邮件、带附件邮件(完整版)
- 学习js第三天小结