CodeForces 635A-Orchestra【图形规划】

A. Orchestra

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

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 ofn violists. Paul really likes violas, so he would like to take a picture including at leastk 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.

Input

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 integersx_i andy_i (1 ≤ x_i ≤ r,1 ≤ y_i ≤ c): the position of thei-th viola. It is guaranteed that no location appears more than once in the input.

Output

Print a single integer — the number of photographs Paul can take which include at leastk violas.

Examples

Input

2 2 1 11 2

Output

4

Input

3 2 3 31 13 12 2

Output

1

Input

3 2 3 21 13 12 2

Output

4

Note

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, the2 × 1 row containing the viola, or the entire string section, for4 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.

解题思路：

题目大意就是r行c列的一个矩阵，将其标记*号，我们会有n次加点的机会，输入一个坐标相应位置会变成#，我们会给出一个K问在这个矩阵里会有多少个包含》=k个#的小矩形，我们可以先将矩阵置零，输入的位置为1。

我们以1，1点为矩形的一个端点，计算（i，j）和（1，1）为对角线的矩形中有多少#。

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;int map[100][100];int an[100][100];int ans[100][100];int main(){int r,c,n,k;while(scanf("%d%d%d%d",&r,&c,&n,&k)!=EOF){memset(map,0,sizeof(map));int i,j;for(i=0;i<n;i++){int x,y;scanf("%d%d",&x,&y);map[x][y]=1;} //int ans=0;int xx,oo;for(i=1;i<=r;i++){for(j=1;j<=c;j++){ans[i][j]=map[i][j]+ans[i][j-1];}}int wc=0;for(i=1;i<=r;i++){for(j=1;j<=c;j++){ans[i][j]+=ans[i-1][j];}}//int xx,oo;for(i=1;i<=r;i++){for(j=1;j<=c;j++){for(xx=i;xx<=r;xx++){for(oo=j;oo<=c;oo++){int ji=ans[xx][oo]+ans[i-1][j-1]-ans[i-1][oo]-ans[xx][j-1];if(ji>=k)wc++;}}}}printf("%d\n",wc);}return 0;}