UVA The die is cast （BFS）

  The die is cast 

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 pixels connected 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 in S, there is a sequence  in S such that a = a₁ and b = a_k , and a_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

       题意：在一个大地图中分布着许多只有*和X的小块，求这些小块中有多少X（其中一个X的上下左右方向如果存在X，那么这个X不进入计数）。最后把所有的小块中的X的值按升序输出。

代码：

#include<iostream>#include<algorithm>#include<stdio.h>#include<string.h>#include<stdlib.h>#include<string>#include<queue>using namespace std;struct node{    int x;    int y;}p[30010];int n,m;int pjx[] = {0,1,0,-1};int pjy[] = {1,0,-1,0};//int jx[] = {0,1,-1,0,1,1,-1,-1};//int jy[] = {1,0,0,-1,1,-1,1,-1};char map[101][101];int v[101][101];int BFS2(int ee){    int cc = 0;    queue<node>q1;    struct node t,f;    memset(v,0,sizeof(v));    for(int i=0;i<ee;i++)    {        if(v[p[i].x][p[i].y] == 0)        {            t.x = p[i].x;            t.y = p[i].y;            v[t.x][t.y] = 1;            map[t.x][t.y] = '.';            q1.push(t);            while(!q1.empty())            {                t = q1.front();                q1.pop();                for(int j=0;j<4;j++)                {                    f.x = t.x + pjx[j];                    f.y = t.y + pjy[j];                    if(f.x>=0 && f.x<n && f.y>=0 && f.y<m && map[f.x][f.y] == 'X')                    {                        q1.push(f);                        map[f.x][f.y] = '.';                        v[f.x][f.y] = 1;                        cc++;                    }                }            }        }    }    return cc;}int BFS(int ii,int jj){    struct node r;    int ee = 0;    int count = 0;    queue<node>q;    memset(v,0,sizeof(v));    node t,f;    t.x = ii;    t.y = jj;    q.push(t);    if(map[t.x][t.y] == 'X')    {        p[ee++] = t;        count++;    }    v[t.x][t.y] = 1;    while(!q.empty())    {        t = q.front();        q.pop();        for(int i=0;i<4;i++)        {            f.x = t.x + pjx[i];            f.y = t.y + pjy[i];            if(f.x>=0 && f.x<n && f.y>=0 && f.y<m && (map[f.x][f.y] == '*' || map[f.x][f.y] == 'X') && v[f.x][f.y] == 0)            {                if(map[f.x][f.y] == 'X')                {                    r.x = f.x;                    r.y = f.y;                    p[ee++] = r;                    count++;                }                else                {                    map[f.x][f.y] = '.';                }                q.push(f);                v[f.x][f.y] = 1;            }        }    }    int cc = BFS2(ee);    return count - cc;}int main(){    int kk = 0;    int a[30100];    while(scanf("%d%d",&m,&n)!=EOF)    {        if(n == 0 && m == 0)        {            break;        }        kk++;        memset(a,0,sizeof(a));        int e = 0;        for(int i=0;i<n;i++)        {            scanf("%s",map[i]);        }        for(int i=0;i<n;i++)        {            for(int j=0;j<m;j++)            {                if(map[i][j] == '*' || map[i][j] == 'X')                {                    int k = BFS(i,j);                    a[e++] = k;                }            }        }        sort(a,a+e);        printf("Throw %d\n",kk);        for(int i=0;i<e-1;i++)        {            printf("%d ",a[i]);        }        printf("%d\n\n",a[e-1]);    }    return 0;}

---

Miguel Revilla
2000-05-22