zoukankan      html  css  js  c++  java
  • codeforces Looksery Cup 2015 D. Haar Features

    The first algorithm for detecting a face on the image working in realtime was developed by Paul Viola and Michael Jones in 2001. A part of the algorithm is a procedure that computes Haar features. As part of this task, we consider a simplified model of this concept.

    Let's consider a rectangular image that is represented with a table of size n × m. The table elements are integers that specify the brightness of each pixel in the image.

    feature also is a rectangular table of size n × m. Each cell of a feature is painted black or white.

    To calculate the value of the given feature at the given image, you must perform the following steps. First the table of the feature is put over the table of the image (without rotations or reflections), thus each pixel is entirely covered with either black or white cell. The valueof a feature in the image is the value of W - B, where W is the total brightness of the pixels in the image, covered with white feature cells, and B is the total brightness of the pixels covered with black feature cells.

    Some examples of the most popular Haar features are given below.

    Your task is to determine the number of operations that are required to calculate the feature by using the so-called prefix rectangles.

    prefix rectangle is any rectangle on the image, the upper left corner of which coincides with the upper left corner of the image.

    You have a variable value, whose value is initially zero. In one operation you can count the sum of pixel values ​​at any prefix rectangle, multiply it by any integer and add to variable value.

    You are given a feature. It is necessary to calculate the minimum number of operations required to calculate the values of this attribute at an arbitrary image. For a better understanding of the statement, read the explanation of the first sample.

    Input

    The first line contains two space-separated integers n and m (1 ≤ n, m ≤ 100) — the number of rows and columns in the feature.

    Next n lines contain the description of the feature. Each line consists of m characters, the j-th character of the i-th line equals to "W", if this element of the feature is white and "B" if it is black.

    Output

    Print a single number — the minimum number of operations that you need to make to calculate the value of the feature.

    Sample test(s)
    input
    6 8
    BBBBBBBB
    BBBBBBBB
    BBBBBBBB
    WWWWWWWW
    WWWWWWWW
    WWWWWWWW
    
    output
    2
    
    input
    3 3
    WBW
    BWW
    WWW
    
    output
    4
    
    input
    3 6
    WWBBWW
    WWBBWW
    WWBBWW
    
    output
    3
    
    input
    4 4
    BBBB
    BBBB
    BBBB
    BBBW
    
    output
    4
    
    Note

    The first sample corresponds to feature B, the one shown in the picture. The value of this feature in an image of size 6 × 8 equals to the difference of the total brightness of the pixels in the lower and upper half of the image. To calculate its value, perform the following twooperations:

    1. add the sum of pixels in the prefix rectangle with the lower right corner in the 6-th row and 8-th column with coefficient 1 to the variable value (the rectangle is indicated by a red frame);
    2. add the number of pixels in the prefix rectangle with the lower right corner in the 3-rd row and 8-th column with coefficient  - 2 and variable value.

    Thus, all the pixels in the lower three rows of the image will be included with factor 1, and all pixels in the upper three rows of the image will be included with factor 1 - 2 =  - 1, as required.

    一开始题意一直看不懂,后来看了题解,发现挺简单的= = 。这题就是从右下角依次向左枚举点,如果是W,就变为1,是B就变为-1。注意更新的时候要从最左上角开始更新。

    #include<iostream>
    #include<stdio.h>
    #include<string.h>
    #include<math.h>
    #include<vector>
    #include<queue>
    #include<stack>
    #include<string>
    #include<algorithm>
    using namespace std;
    char s[105][105];
    int map[105][106];
    int main()
    {
    	int n,m,i,j,c,ans,r,num;
    	while(scanf("%d%d",&n,&m)!=EOF)
    	{
    		for(i=0;i<n;i++){
    			scanf("%s",s[i]);
    		}
    		num=0;
    		memset(map,0,sizeof(map));
    		for(i=n-1;i>=0;i--){
    			for(j=m-1;j>=0;j--){
    				if(s[i][j]=='W'){
    					if(map[i][j]!=1){
    						num++;
    						ans=1-map[i][j];
    						for(r=i;r>=0;r--){
    							for(c=j;c>=0;c--){
    								map[r][c]+=ans;
    							}
    						}
    					}	
    				}
    				else if(s[i][j]=='B'){
    					if(map[i][j]!=-1){
    						num++;
    						ans=map[i][j]+1;
    						for(r=i;r>=0;r--){
    							for(c=j;c>=0;c--){
    								map[r][c]-=ans;
    							}
    						}
    					}
    				}
    			}
    		}
    		printf("%d
    ",num);
    	}
    	return 0;
    }


  • 相关阅读:
    CQOI2016 不同的最小割 (最小割树模板)(等价流树的Gusfield构造算法)
    CSP2019 D2T2 划分 (单调队列DP)
    Android弱网测试中关于网络检测的一些借鉴方法
    IOS的Crash情况在Crashlytics平台上统计解决方案的一点遗憾(截止到2015年6月14日)
    关于selenium2(webdriver)自动化测试过程中标签页面或者窗口切换的处理解决方案
    针对电信乌龙事件的深度测试: 广州电信错误将深圳地区189的号码在3G升级4G申请时从广州网厅发货,造成深圳用户收到4G卡后无法激活,深圳电信找不到订单
    spring mvc + freemarker优雅的实现邮件定时发送
    使用phantomjs实现highcharts等报表通过邮件发送(本文仅提供完整解决方案和实现思路,完全照搬不去整理代码无法马上得到效果)
    模拟http或https请求,实现ssl下的bugzilla登录、新增BUG,保持会话以及处理token
    xmlrpc实现bugzilla api调用(无会话保持功能,单一接口请求)
  • 原文地址:https://www.cnblogs.com/herumw/p/9464762.html
Copyright © 2011-2022 走看看