UVA 657 The die is cast ——bfs+dfs
2014-08-18 09:16
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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
inS 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.
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 valuessatisfy
.
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
Miguel Revilla
2000-05-22
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <algorithm>
using namespace std;
char mp[55][55];//存储地图
int vis[55][55];//标记是否被访问过
int s[110];//答案存储在这个数组中
struct node
{
int x;
int y;
} ls[3601],t,f;//bfs的队列
int bhx[] = {-1,0,1,0};
int bhy[] = {0,1,0,-1};//四个方向
int num = 0;//记录X的个数
int js;
int m,n;
void dfs(int x,int y)
{
int i,a,b;
vis[x][y] = 1;//标记X
for(i = 0; i < 4; i++)
{
a = x + bhx[i];
b = y + bhy[i];
if(a >= 0 && a < n&& b >= 0 && b < m && mp[a][b] == 'X' && !vis[a][b])
{
dfs(a,b);
}
}
}
void bfs(int x,int y)
{
int tp = 0,tl = 0,i;
t.x = x;
t.y = y;
ls[tp++] = t;//第一个元素进队列
vis[x][y] = 1;//标记第一个元素
while(tl < tp)
{
t = ls[tl++];
if(mp[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);//开始遍历
num++;//每调用一次dfs,就是发现了一个或是一群X
}
for(i = 0; i < 4; i++)
{
f.x = t.x + bhx[i];
f.y = t.y + bhy[i];
if(f.x >= 0 &&f.x < n &&f.y >= 0 && f.y < m && !vis[f.x][f.y] && mp[f.x][f.y] != '.')//在*群里搜索,找X
{
ls[tp++] = f;
if(mp[f.x][f.y] != 'X')//是X的话就不标记,X在dfs标记
vis[f.x][f.y] = 1;
}
}
}
if(num != 0)//如果num是0的话,说明没发现*群里有X,就不统计
{
s[js++] = num;
num = 0;
}
}
int main()
{
int i,j;
int c = 0;
while(scanf("%d%d",&m,&n),n||m)
{
c++;
js = 0;
memset(vis,0,sizeof(vis));//vis标记清空
for(i = 0; i < n; i++)
{
scanf("%*c%s",mp[i]);
}//把地图读入
for(i = 0; i < n; i++)
{
for(j = 0; j <m; j++)
{
if((mp[i][j] == '*'|| mp[i][j] == 'X') && !vis[i][j])//循环访问地图,合适的进行bfs搜索
{
bfs(i,j);
}
}
}
sort(s,s+js);
printf("Throw %d\n",c);
for(i = 0; i < js; i++)
{
printf("%d",s[i]);
if(i < js - 1)
printf(" ");
}
printf("\n\n");
}
return 0;
}
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