Uva 657 The die is cast

The die is cast

Time Limit: 3000MS Memory Limit: Unknown 64bit IO Format: %lld & %llu

InterGames is a high-tech startup company that specializes in developing technology that allows usersto play games over the Internet. A market analysis has alerted them to the fact that games of chanceare pretty popular among their potential customers. Be it Monopoly, ludo or backgammon, most ofthese 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 theresult into the computer, since cheating would be way to easy. So, instead, InterGames has decided tosupply their users with a camera that takes a picture of the thrown dice, analyzes the picture and thentransmits the outcome of the throw automatically.

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

We make the following assumptions about the input images. The imagescontain only three different pixel values: for the background, the diceand the dots on the dice. We consider two pixels connected if they sharean edge — meeting at a corner is not enough. In the figure, pixels A andB are connected, but B and C are not.

A set S of pixels is connected if for every pair (a, b) of pixels in S, thereis a sequence a1, a2, . . . , ak in S such that a = a1 and b = ak, and ai andai+1 are connected for 1 ≤ i < k.

We consider all maximally connected sets consisting solely of nonbackgroundpixels to be dice. ‘Maximally connected’ means that you cannotadd any other non-background pixels to the set without making it dis-connected. Likewise weconsider 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 linecontaining two numbers w and h, the width and height of the picture, respectively. These values satisfy5 ≤ w, h ≤ 50.

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 willcontain 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 thepicture, 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 1

1 2 2 4

开始以为*会把X包一圈，但好像并非如此；

搜索时，遇到X，要把周围的X打上标记，防止重复计点数；

代码写得比较丑；

#include<iostream>#include<cstdlib>#include<string>#include<algorithm>#include<cstdio>#include<cmath>#include<cstring>#include<stack>#include<queue>#include<iomanip>#include<map>#include<set>#include<functional>#define pi 3.14159265358979323846using namespace std;int dx[]={1,0,0,-1};int dy[]={0,1,-1,0};char maze[55][55];bool vis[55][55];bool flag[55][55];int num[1250];int w,h;void dfs(int x,int y,int tol){    if(!vis[x][y]&&maze[x][y]!='.')        vis[x][y]=1;    else return ;    if(maze[x][y]=='X')    {        if(flag[x][y]==0)        {            num[tol]++;            flag[x][y]=1;        }        for(int i=0;i<4;++i)        {            if(x+dx[i]<h&&x+dx[i]>=0&&y+dy[i]<w&&y+dy[i]>=0&&!vis[x+dx[i]][y+dy[i]]&&maze[x+dx[i]][y+dy[i]]=='X')            {                flag[x+dx[i]][y+dy[i]]=1;            }        }        for(int i=0;i<4;++i)        {            if(x+dx[i]<h&&x+dx[i]>=0&&y+dy[i]<w&&y+dy[i]>=0&&!vis[x+dx[i]][y+dy[i]]&&maze[x+dx[i]][y+dy[i]]=='X')            {                dfs(x+dx[i],y+dy[i],tol);            }        }    }        for(int i=0;i<4;++i)        {            if(x+dx[i]<h&&x+dx[i]>=0&&y+dy[i]<w&&y+dy[i]>=0&&!vis[x+dx[i]][y+dy[i]])            {                dfs(x+dx[i],y+dy[i],tol);            }        }    return ;}int main(){    int cnt=0;    while(scanf("%d %d",&w,&h)!=EOF&&(w||h))    {        for(int i=0;i<h;++i)        {                scanf(" %s",&maze[i]);        }        memset(vis,0,sizeof(vis));        memset(num,0,sizeof(num));        memset(flag,0,sizeof(flag));        int tol=0;        printf("Throw %d\n",++cnt);        for(int i=0;i<h;++i)        {            for(int j=0;j<w;++j)            {                if(!vis[i][j]&&maze[i][j]!='.')                {                    dfs(i,j,tol++);                }            }        }        sort(num,num+tol);        for(int i=0;i<tol;++i)        {            if(i==0)                printf("%d",num[i]);            else printf(" %d",num[i]);        }        printf("\n\n");    }    return 0;}