[657] The die is cast UVA



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 different 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 a1, a2, ..., ak in S such that a = a1 and b = ak , and a_i and a_i+1are connected for 1 <= i < k. 

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 5 <= 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 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 1
1 2 2 4

本题主要运用了BFS和DFS的思想，应注意以下几点：

（一）输入行和列时应弄清楚先输列数再输行数

（二）答案应从小到大排序

（三）输出时应多输一行空行

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
char ma[60][60];//记录地图
int vis[60][60];//标记是否访问
int an[120];//记录最终答案
struct node
{
    int x,y;
} q[3600],f,t;
int mv[4][2]= {{1,0},{0,1},{0,-1},{-1,0}};//方向
int num=0;//记录每组‘*’中'X'的个数；
int gs=0;//记录‘*’堆的组数
int n,m;
void qsort(int *a,int star,int fin)//快排
{
    int key=a[star],i=star,j=fin;
    if(star>fin)
       return;
    while(i<j)
    {
        while(i<j&&a[j]>=key) j--;
          a[i]=a[j];
          while(i<j&&a[i]<=key) i++;
          a[j]=a[i];
    }
    a[i]=key;
    qsort(a,star,i-1);
    qsort(a,i+1,fin);
}
void dfs(int x,int y)
{
    int i;
    struct node f1;
    vis[x][y]=1;
    for(i=0;i<4;i++)
    {
        f1.x=x+mv[i][0];
        f1.y=y+mv[i][1];
        if(f1.x>=0&&f1.x<n&&f1.y>=0&&f1.y<m&&!vis[f1.x][f1.y]&&ma[f1.x][f1.y]=='X')
        {
            dfs(f1.x,f1.y);
        }
    }
}
void bfs(int x,int y)
{
    f.x=x;
    f.y=y;
    int jin=0,chu=0;
    q[jin++]=f;
    while(chu<jin)
    {
        t=q[chu++];
        if(ma[t.x][t.y]=='X'&&!vis[t.x][t.y])//如果某个X未被访问过
        {
            dfs(t.x,t.y);
            num++;//每调用一次dfs，就是发现了一个或是一群X  
        }
        for(int i=0; i<4; i++)
        {
            f.x=t.x+mv[i][0];
            f.y=t.y+mv[i][1];
            if(f.x>=0&&f.x<n&&f.y>=0&&f.y<m&&!vis[f.x][f.y]&&ma[f.x][f.y]!='.')//在*群里搜索，找X
            {
                if(ma[f.x][f.y]!='X')//是X的话就不标记，X在dfs标记  
                    vis[f.x][f.y]=1;
                q[jin++]=f;
            }
        }
    }
    if(num!=0)//如果num是0的话，说明没发现*群里有X，就不统计
    {
        an[gs++]=num;
        num=0;
    }
}
int main()
{
    int i,j;
    int time=0;
    while(~scanf("%d %d",&m,&n)&&n!=0&&m!=0)
    {
        time++;
        getchar();
        memset(vis,0,sizeof(vis));
        for(i=0; i<n; i++)
            scanf("%s",ma[i]);
        for(i=0; i<n; i++)
        {
            for(j=0; j<m; j++)
            {
                if((ma[i][j]=='*'||ma[i][j]=='X')&&!vis[i][j])//循环访问地图，合适的进行bfs搜索  
                {
                    bfs(i,j);
                }
            }
        }
        qsort(an,0,gs-1);
        printf("Throw %d\n",time);
        for(i=0;i<gs;i++)
        {
            if(i!=gs-1)
                printf("%d ",an[i]);
            else
                printf("%d\n\n",an[i]);
        }
        gs=0;
    }
    return 0;
}

转载于:https://www.cnblogs.com/jiangyongy/p/3971668.html