swystoj(OpenCV(0272))

OpenCV (Open Source Computer Vision) is a library of programming functions for real time computer vision. OpenCV is released under a BSD license, it is free for both academic and commercial use. Example applications of the OpenCV library are Human-Computer Interaction (HCI); Object Identification, Segmentation and Recognition; Face Recognition; Gesture Recognition; Camera and Motion Tracking, Ego Motion, Motion Understanding; Structure From Motion (SFM); Stereo and Multi-Camera Calibration and Depth Computation; Mobile Robotics. Recently ,Phillip has learnt about how to use OpenCV to solve some image problems. Doctor Jiang wants him find the moving items from the given video files. Since using OpenCV can easily deal with such problems, Phillip wants to use his own code to make it out. As we kwon images are saved as a lot of pixels in the computer. The algorithm that Phillip use is below: First, we find the different pixels between current image and the previous one image(in this problem we only care the different pixels) . Second, we record the maximum different number . Description The first line are two integers r (row)and c(column) which describe the size of one image. (1 < r, c < 200 ); The second Line is n which is the total number of images (1 < n < 100) Then n*r lines follow, each line contain exactly c numbers which are pixels of the image in the range of 0 to 255. Input One line with the maximum differenct number you have found. Output 4 4 3 255 0 0 255 223 221 220 190 10 11 12 13 0 0 0 0 255 1 1 255 245 221 220 100 11 11 0 13 1 1 1 1 255 0 0 255 245 221 220 100 11 11 0 13 1 1 1 1 Sample Input 10 #include <stdio.h> #include <string.h> #include <algorithm> #include<iostream> using namespace std; struct node { int a[201][201]; }b[101]; int main() { int n, m, l; while (cin>>n>>m>>l) { int ans = 0; for (int i = 0; i < l; i++) { int count = 0; for (int j = 0; j < n; j++) { for (int k = 0; k < m; k++) { cin >> b[i].a[j][k]; if (i&&b[i].a[j][k] != b[i - 1].a[j][k]) count++;//在输入是判断，减少时间。 } } if (count > ans) ans = count; } cout << ans << endl; } return 0; }