uva657 The die is cast（搜索）

思路：两次DFS...先读懂题意..

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
char map[55][55];
int num[1000],t,w,h;
int dir[4][2]={{1,0},{0,-1},{-1,0},{0,1}};
void dfs1(int x,int y)
{
    int xx,yy,i;
    map[x][y]='*';
    for(i=0;i<4;i++)
    {
        xx=x+dir[i][0];
        yy=y+dir[i][1];
        if(xx>=h||yy>=w||xx<0||yy<0||map[xx][yy]=='*')
           continue;
        if(map[xx][yy]=='X')
           dfs1(xx,yy);
    }
}
void dfs2(int x,int y)
{
    int xx,yy,i;
    map[x][y]='.';
    for(i=0;i<4;i++)
    {
        xx=x+dir[i][0];
        yy=y+dir[i][1];
        if(xx>=h||yy>=w||xx<0||yy<0||map[xx][yy]=='.')
           continue;
        if(map[xx][yy]=='X')
        {
            dfs1(xx,yy);
            num[t]++;
        }
        if(map[xx][yy]=='*')
            dfs2(xx,yy);
    }
}
int main()
{
    int i,j,n=0;
    while(~scanf("%d%d",&w,&h)&&w&&h)
    {
        memset(num,0,sizeof(num)); /*记得每次用过后都要清0*/
        memset(map,'.',sizeof(map));
        t=0;
        for(i=0;i<h;i++)
        {
            getchar();
            for(j=0;j<w;j++)
              scanf("%c",&map[i][j]);
        }
        for(i=0;i<h;i++)
          for(j=0;j<w;j++)
            if(map[i][j]=='*')
            {
                dfs2(i,j);
                t++;
            }
        sort(num,num+t);
        printf("Throw %d\n",++n);
        for(i=0;i<t-1;i++)
          printf("%d ",num[i]);
        printf("%d\n",num[i]);
        printf("\n");
    }
    return 0;
}

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 ai and
ai+1 are connected for 1 ≤ i < k.
We consider all maximally connected sets consisting solely of nonbackground
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