UVA - 657
| Time Limit: 3000MS | Memory Limit: Unknown | 64bit IO Format: %lld & %llu |
Description
| 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 = ak , and ai and ai+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.
.
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.
Print a blank line after each test case.
30 15 .............................. .............................. ...............*.............. ...*****......****............ ...*X***.....**X***........... ...*****....***X**............ ...***X*.....****............. ...*****.......*.............. .............................. ........***........******..... .......**X****.....*X**X*..... ......*******......******..... .....****X**.......*X**X*..... ........***........******..... .............................. 0 0
Throw 1 1 2 2 4
Source
题意：大概是去找每个dice中的连在一起的dots的个数（每两个dots连着的判定条件是他们的边相连）， 输出时按从小到大输出， 还有注意每组数据之后要有空行
思路：两次DFS，先找到dice，再去找dice里面连在一起的dots的个数
AC代码：
#include <cstdio>
#include <cstring>
#include <cmath>
#include <string>
#include <algorithm>
using namespace std;
int dir[4][2]={{1,0},{0,-1},{-1,0},{0,1}};
const int maxn = 55;
char map[maxn][maxn];
int vis[maxn][maxn];
int ans[1300];
int num, w, h;
void dfs2(int x, int y)
{
if(map[x][y] == '*' || map[x][y] == '.' || vis[x][y]) return;
vis[x][y] = 1;
for(int i=0; i<4; i++)
{
int xx=x+dir[i][0];
int yy=y+dir[i][1];
if(xx>=h||yy>=w||xx<0||yy<0||map[xx][yy]=='*') //没判断是否溢出数组，RE一次
continue;
dfs2(xx, yy);
}
}
void dfs(int x, int y)
{
if(map[x][y] == '.' || vis[x][y]) return;
if(!vis[x][y] && map[x][y] == 'X') //找到dice中的其中一个dots然后再进行二次DFS
{
ans[num]++;
dfs2(x, y);
}
vis[x][y] = 1;
for(int i=0; i<4; i++)
{
int xx=x+dir[i][0];
int yy=y+dir[i][1];
if(xx>=h||yy>=w||xx<0||yy<0||map[xx][yy]=='.') //没判断是否溢出数组，RE一次
continue;
dfs(xx,yy);
}
}
int main()
{
int cas = 1;
while(scanf("%d %d", &w, &h), w || h)
{
char a[55];
for(int i=0; i<h; i++)
{
scanf("%s", a);
for(int j=0; j<w; j++)
{
map[i][j] = a[j];
}
}
memset(vis, 0, sizeof(vis));
memset(ans, 0, sizeof(ans));
num = 0;
for(int i=0; i<h; i++)
{
for(int j=0; j<w; j++)
{
if(!vis[i][j] && map[i][j] == '*') //找到dice然后进行第一次DFS
{ dfs(i, j); num++; }
}
}
sort(ans, ans+num);
printf("Throw %d\n", cas++);
for(int i=0; i<num-1; i++)
{
if(ans[i]) printf("%d ", ans[i]);
}
printf("%d\n\n", ans[num-1]);
}
return 0;
}
UVA - 657 - The die is cast （两次DFS！！）
原文地址：http://blog.csdn.net/u014355480/article/details/41970611
