Va 657-The die is cast (dual BFS)

Va 657-The die is cast (dual BFS)

InterGames is a high-tech startup company that specializes in developing technology that allows users to play games over the Internet. A market analysis has alerted them to the fact that games of chance are pretty popular among their potential MERs. be it Monopoly, ludo or backgammon, most of these games involve throwing dice at some stage of the game.

Of course, it wocould be unreasonable if players were allowed to throw their dice and then enter the result into the computer, since cheating wocould be way to easy. so, instead, InterGames has decided to supply their users with a camera that takes a picture of the thrown dice, analyzes the picture and then transmits the outcome of the throw automatically.

For this they desperately need a program that, given an image containing several dice, determines the numbers of dots on the dice.

We make the following assumptions about the input images. The images contain only three dif-ferent pixel values: for the background, the dice and the dots on the dice. We consider twpixo elsConnectedIf they share an edge-meeting at a corner is not enough. In the figure, pixels A and B are connected, but B and C are not.

A setSOf pixels is connected if for every pair (A,B) Of pixels inS, There is a sequence inSSuch thatA=A1 andB=AK, AndAIAndAI+ 1 are connected.

We consider all maximally connected sets consisting solely of non-background pixels to be dice. 'maximally connected' means that you cannot add any other non-background pixels to the set without making it dis-connected. likewise we consider every maximal connected set of dot pixels to form a dot.

Input The input consists of pictures of several dice throws. Each picture description starts with a line containing two numbers w and h, the width and height of the picture, respectively. These values satisfy.

The followingHLines containWCharacters each. The characters can be :''.''For a background pixel ,''*''For a pixel of a die, and''X''For a pixel of a die's dot.

Dice may have different sizes and not be entirely square due to optical distortion. The picture will contain at least die one, and the numbers of dots per die is between 1 and 6, random Sive.

The input is terminated by a picture startingW=H= 0, which shocould not be processed.

Output For each throw of dice, first output its number. Then output the number of dots on the dice in the picture, sorted in increasing order.

Print a blank line after each test case.

Sample Input

30 15...........................................................................*.................*****......****...............*X***.....**X***..............*****....***X**...............***X*.....****................*****.......*....................................................***........******............**X****.....*X**X*...........*******......******..........****X**.......*X**X*.............***........******...................................0 0

Sample Output

Throw 11 2 2 4

Finally, I realized that it would not be a DFS problem. This question was answered six times after BFSwa. Let's talk about the idea and use the for loop traversal matrix, if you encounter * Go To BFS search * (Set * .), Search for X in BFS (set the searched X *)

Note that the xxxx answer is 0 for special data.

#include 
 
  #include 
  
   #include 
   
    #include 
    
     #include 
     
      #include using namespace std;typedef struct node{int x,y;node (int a,int b){x=a;y=b;}};int dir[4][2] = {{0, 1}, {0, -1}, {1, 0}, { -1, 0}};char ma[60][60];int cnt,m, n, ans[1000];void bfs2(int x, int y){  ma[x][y]='*';  queue 
      
        S; S.push(node(x,y)); while(!S.empty()) { node v=S.front();S.pop();for (int i = 0; i < 4; i++){int tx = v.x + dir[i][0];int ty = v.y + dir[i][1];if (tx>=0&&tx
       
        =0&&ty
        
          Q;ma[x][y]='.';Q.push(node(x, y));while (!Q.empty()){node v = Q.front();Q.pop();if (ma[v.x][v.y] == 'X'){bfs2(v.x, v.y);ans[cnt]++;}else if(ma[v.x][v.y]=='*'){ma[v.x][v.y]='.';}for (int i = 0; i < 4; i++){int tx = v.x + dir[i][0];int ty = v.y + dir[i][1];if (ma[tx][ty]!='.'&&tx>=0&&tx
         
          =0&&ty
          
           > n >> m){if (!m && !n){break;}cnt=0;getchar();memset(ans, 0, sizeof(ans));memset(ma,'.',sizeof(ma));for (i = 0; i < m; i++){cin >> ma[i];}for (i = 0; i < m; i++)for (int j = 0; j < n; j++)if (ma[i][j]=='*'){bfs1(i, j);}sort(ans,ans+cnt);cout<<"Throw "<