[POJ]1111 Image Perimeters

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 2

XX

XX

6 4 2 3

.XXX

.XXX

.XXX

…X

..X.

X…

5 6 1 3

.XXXX.

X….X

..XX.X

.X…X

..XXX.

7 7 2 6

XXXXXXX

XX…XX

X..X..X

X..X…

X..X..X

X…..X

XXXXXXX

7 7 4 4

XXXXXXX

XX…XX

X..X..X X..X…

X..X..X

X…..X

XXXXXXX

0 0 0 0

Sample Output

8

18

40

48

8

Source

Mid-Central USA 2001

一开始周长统计想了好久，后来发现旁边是空格就行。

还是很像swim..嗯

#include<iostream>
#include<cstring>
#include<queue>
#define MAXN 25
using namespace std;

char a[MAXN][MAXN];
bool vis[MAXN][MAXN];
bool b[MAXN][MAXN][5];
int m,n,sx,sy;

int dx[8]= {0,1,1,1,0,-1,-1,-1};
int dy[8]= {1,1,0,-1,-1,-1,0,1};

inline bool pd(int x,int y) {
    if(x<1||x>m||y<1||y>n) return false;
    return true;
}

void dfs(int x,int y) {
    if(!pd(x,y)) return;
    a[x][y]='O';
    for(int i=0; i<=7; i++) {
        int nx=x+dx[i],ny=y+dy[i];
        if(pd(nx,ny)) {
            if(!vis[nx][ny]&&a[nx][ny]=='X') {
                vis[nx][ny]=1;
                dfs(nx,ny);

            }
        }
    }
}

int calc() {
    int i,j;
    int cnt=0;
    for(i=1; i<=m; i++) {
        for(j=1; j<=n; j++) {
            if(a[i][j]=='O') {
                if(a[i+1][j]=='.'||a[i+1][j]==0) cnt++;
                if(a[i-1][j]=='.'||a[i-1][j]==0) cnt++;
                if(a[i][j+1]=='.'||a[i][j+1]==0) cnt++;
                if(a[i][j-1]=='.'||a[i][j-1]==0) cnt++;
            }
        }
    }
    return cnt;
}

int main() {
    while(cin>>m>>n>>sx>>sy) {
        if(m==0&&n==0&&sx==0&&sy==0) return 0;
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
        memset(vis,0,sizeof(vis));
        int i,j;
        for(i=1; i<=m; i++) {
            for(j=1; j<=n; j++) {
                cin>>a[i][j];
            }
        }
        dfs(sx,sy);
        int ans=calc();
        cout<<ans<<endl;
    }
    return 0;
}