Nyoj 92 useful areas of images

Useful areas of imagesTime Limit: 3000 MS | memory limit: 65535 kb difficulty: 4

Description

"ACKing" encountered a problem when he was working on an image processing project. He needed to extract images within the area of a black line in the image, now, please help him complete the first step and turn all the areas outside the black line into black.

Figure 1 Figure 2

It is known that there will be no crossover everywhere in the black line (2), and, except for the black line points, there is no pure black in the image (that is, the pixel is 0 points ).

 

Input

Number of groups of the first input test data N (0 <n <= 6)
The first row of each group of test data is two integers W, and the H table represents the width and height of the image (3 <= W <= 1440,3 <= H <= 960)
In the subsequent row H, each row has W positive integers, indicating the pixel value of the point. (Pixels are between 0 and 255. 0 indicates black and 255 indicates white)

Output

Output the pixel values of each point in the image after the black area is black in the form of a matrix.

Sample Input

15 5100 253 214 146 120123 0 0 0 054 0 33 47 0255 0 0 78 014 11 0 0 0

Sample output

0 0 0 0 00 0 0 0 00 0 33 47 00 0 0 78 00 0 0 0 0

Source

[Zhang yuncong] original

Uploaded

Zhang yuncong

Problem solving: It's amazing that BFS can do this kind of thing ............. Outside the world ........

 1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cstdlib> 5 #include <vector> 6 #include <climits> 7 #include <algorithm> 8 #include <cmath> 9 #include <queue>10 #define LL long long11 using namespace std;12 int rows,cols,mp[1000][1500];13 queue<int>q;14 void bfs(int x,int y) {15     const int dir[4][2] = {0,-1,0,1,-1,0,1,0};16     int i,j,tx,ty,u,v;17     q.push(x);18     q.push(y);19     while(!q.empty()) {20         tx = q.front();21         q.pop();22         ty = q.front();23         q.pop();24         for(i = 0; i < 4; i++) {25             u = tx+dir[i][0];26             v = ty+dir[i][1];27             if(!mp[u][v]) continue;28             if(u < 0 || v < 0 || u >= 999 || v >= 1499) continue;29             mp[u][v] = 0;30             q.push(u);31             q.push(v);32         }33     }34 }35 int main() {36     int ks,i,j;37     scanf("%d",&ks);38     while(ks--) {39         scanf("%d %d",&cols,&rows);40         for(i = 0; i < 1000; i++)41             for(j = 0; j < 1500; j++)42                 mp[i][j] = 1;43         for(i = 1; i <= rows; i++) {44             for(j = 1; j <= cols; j++)45                 scanf("%d",mp[i]+j);46         }47         while(!q.empty()) q.pop();48         bfs(0,0);49         for(i = 1; i <= rows; i++) {50             for(j = 1; j <= cols-1; j++)51                 printf("%d ",mp[i][j]);52             printf("%d\n",mp[i][j]);53         }54     }55     return 0;56 }

View code

Why do we need to add a circle of 1? Hey ..... Because the search starts from the upper left corner. If it is all 0, it may not be on the right .. Then the search will fail, resulting in wa ..... The ring is added to ensure that the outside of the Zero ring is a continuous one.