poj1111之DFS

Image Perimeters

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 9120 Accepted: 5395

Description

Technicians in a pathology lab analyze digitized images of slides. Objects on a slide are selected for analysis by a mouse click on the object. The perimeter of the boundary of an object is one useful measure. Your task is to determine this perimeter for selected objects. 

The digitized slides will be represented by a rectangular grid of periods, '.', indicating empty space, and the capital letter 'X', indicating part of an object. Simple examples are 

XX   Grid 1       .XXX   Grid 2 XX                .XXX                   .XXX                   ...X                   ..X.                   X... 

An X in a grid square indicates that the entire grid square, including its boundaries, lies in some object. The X in the center of the grid below is adjacent to the X in any of the 8 positions around it. The grid squares for any two adjacent X's overlap on an edge or corner, so they are connected. 

XXX XXX    Central X and adjacent X's XXX 

An object consists of the grid squares of all X's that can be linked to one another through a sequence of adjacent X's. In Grid 1, the whole grid is filled by one object. In Grid 2 there are two objects. One object contains only the lower left grid square. The remaining X's belong to the other object. 

The technician will always click on an X, selecting the object containing that X. The coordinates of the click are recorded. Rows and columns are numbered starting from 1 in the upper left hand corner. The technician could select the object in Grid 1 by clicking on row 2 and column 2. The larger object in Grid 2 could be selected by clicking on row 2, column 3. The click could not be on row 4, column 3. 

One useful statistic is the perimeter of the object. Assume each X corresponds to a square one unit on each side. Hence the object in Grid 1 has perimeter 8 (2 on each of four sides). The perimeter for the larger object in Grid 2 is illustrated in the figure at the left. The length is 18. 

Objects will not contain any totally enclosed holes, so the leftmost grid patterns shown below could NOT appear. The variations on the right could appear: 

Impossible   Possible XXXX         XXXX   XXXX   XXXX X..X         XXXX   X...   X... XX.X         XXXX   XX.X   XX.X XXXX         XXXX   XXXX   XX.X .....        .....  .....  ..... ..X..        ..X..  ..X..  ..X.. .X.X.        .XXX.  .X...  ..... ..X..        ..X..  ..X..  ..X.. .....        .....  .....  ..... 

Input

The input will contain one or more grids. Each grid is preceded by a line containing the number of rows and columns in the grid and the row and column of the mouse click. All numbers are in the range 1-20. The rows of the grid follow, starting on the next line, consisting of '.' and 'X' characters. 

The end of the input is indicated by a line containing four zeros. The numbers on any one line are separated by blanks. The grid rows contain no blanks. 

Output

For each grid in the input, the output contains a single line with the perimeter of the specified object.

Sample Input

2 2 2 2XXXX6 4 2 3.XXX.XXX.XXX...X..X.X...5 6 1 3.XXXX.X....X..XX.X.X...X..XXX.7 7 2 6XXXXXXXXX...XXX..X..XX..X...X..X..XX.....XXXXXXXX7 7 4 4XXXXXXXXX...XXX..X..XX..X...X..X..XX.....XXXXXXXX0 0 0 0

Sample Output

81840488

Source

Mid-Central USA 2001

题目大意：求出要围成给出的坐标所在点的八联通块所需的物体个数。简单来说，就是对于这个八联通块的每一个‘X’点，考虑其上下左右四个方向是否为“X”，不是就加一，最后得出的结果就是答案。

题目分析：题目很简单，就是题目看不懂，简单的dfs就可以做出来，最开始没有对map初始化导致wa了一次。

代码：

#include <iostream>#include <string.h>using namespace std;#define size 50int ans,w,h,x,y;char map[size][size];int visit[size][size],to[4][2]={1,0,-1,0,0,1,0,-1};void dfs(int x,int y){    for(int i=0;i<4;i++){        if(map[x+to[i][0]][y+to[i][1]]!='X'){            ans++;        }    }    if(map[x][y]=='X'){        visit[x][y]=1;        for(int i=-1;i<2;i++){            for(int j=-1;j<2;j++){                if(x+i>=1&&x+i<=h&&y+j>=1&&y+j<=w&&!visit[x+i][y+j]&&map[x+i][y+j]=='X'){                    dfs(x+i,y+j);                }            }        }    }}int main(){    while(cin>>h>>w>>x>>y&&(w||h||x||y)){        ans=0;        memset(map,'.',sizeof(map));        for(int i=1;i<=h;i++){            for(int j=1;j<=w;j++){                cin>>map[i][j];                visit[i][j]=0;            }        }        dfs(x,y);        cout<<ans<<endl;    }    return 0;}