UVA657

UVA657（题目pdf有些乱码...别在意）

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 inS, there

is a sequence a1; a2; : : : ; ak in S such thata = 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

这是油田的那道题的升级版，需要每个在点阵里断开的’*’围出来的区域里的上下左右不相互连通的X的数量，相信油田的题大家都很熟悉，这个题目只需要在油田的基础上再套用一重DFS即可。

 

下面给出AC代码：

//双重DFS#include<iostream>#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;#define MAX 55int w,h,t;char Map[MAX][MAX];int num[1000];int dir[4][2]={{1,0},{-1,0},{0,1},{0,-1}};void DFS1(int x,int y){Map[x][y]='*';int xx,yy;for(int i=0;i<4;i++){xx=x+dir[i][0];yy=y+dir[i][1];if(xx>0&&xx<=h&&yy>0&&yy<=w){if(Map[xx][yy]=='X') DFS1(xx,yy);}}}void DFS2(int x,int y){int xx,yy;Map[x][y]='.';for(int i=0;i<4;i++){xx=x+dir[i][0];yy=y+dir[i][1];if(xx>0&&xx<=h&&yy>0&&yy<=w){if(Map[xx][yy]=='X'){num[t]++;DFS1(xx,yy);}if(Map[xx][yy]=='*'){DFS2(xx,yy);}}}}int main(){int n=0;while(scanf("%d%d",&w,&h)!=EOF){if(w==0&&h==0) break;memset(Map,'.',sizeof(Map));memset(num,0,sizeof(num));t=0;for(int i=1;i<=h;i++)for(int j=1;j<=w;j++)    cin>>Map[i][j];for(int i=1;i<=h;i++)for(int j=1;j<=w;j++){if(Map[i][j]=='*'){DFS2(i,j);t++;}}sort(num,num+t);printf("Throw %d\n",++n);printf("%d",num[0]);if(t>1)for(int i=1;i<t;i++)printf(" %d",num[i]);printf("\n\n");}return 0;}