The die is cast



The die is cast

Description

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 customers. Be it Monopoly, ludo or backgammon, most of these games involve throwing dice at some stage of the game.

Of course, it would be unreasonable if players were allowed to throw their dice and then enter the result into the computer, since cheating would 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 two pixelsconnected if 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 set S of pixels is connected if for every pair (a,b) of pixels inS, there is a sequence inS such that a = a₁ andb = a_k , anda_i and a_i+1 are connected for.

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 following h lines contain w characters 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 one die, and the numbers of dots per die is between 1 and 6, inclusive.

The input is terminated by a picture starting with w = h = 0, which should 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

大意：

所给的图形中’.‘是背景，‘*’是骰子的骰子的边界，X是骰子的点数，如果多个X相邻（上，下，左，右）记作一点；

求出所有骰子的点数，按升序输出；

要点：

30 15 列 行；

不存在只有X组成的骰子 ；

两重DFS第一重找*，第二重找X，找到X要标记为*，返回时要再次在刚才找到的X（此时已为*）处DFS找*；

代码：

#include <cstdio>#include <algorithm>#include <string.h>using namespace std;char mn[55][55];int dir[4][2] = { { -1, 0 }, { 1, 0 }, { 0, -1 }, { 0, 1 }};int m, n, sum, ans[100];void dfs2(int x, int y){mn[x][y] = '*';for (int i = 0; i < 4; i++){int a = x + dir[i][0];int b = y + dir[i][1];if (!(a >= m || b >= n || a < 0 || b < 0)){if (mn[a][b] == 'X')dfs2(a, b);}}}void dfs1(int x, int y){mn[x][y] = '.';for (int i = 0; i < 4; i++){int a = x + dir[i][0];int b = y + dir[i][1];if (!(a >= m || b >= n || a < 0 || b < 0)){if (mn[a][b] == 'X'){     //注意两个dfs的调用顺序dfs2(a, b);sum++;}if (mn[a][b] == '*'){dfs1(a, b);}}}}int main(){int t = 1;while (scanf("%d%d", &n, &m) && m != 0){for (int i = 0; i < m; i++){scanf("%s", mn[i]);}int num = 0;for (int i = 0; i < m; i++){for (int j = 0; j < n; j++)if (mn[i][j] == '*'){sum = 0;dfs1(i, j);ans[num++] = sum;}}printf("Throw %d\n", t++);sort(ans, ans + num);int l;for (l = 0; l < num - 1; l++)printf("%d ", ans[l]);printf("%d\n\n", ans[l]);memset(ans, 0, sizeof(ans));}return 0;}