Poj 1111 Image Perimeters-优快云博客

题意：理解为搜索一幅图中‘X’相连组成图形的周长（从输入中给出的点开始搜索），相连的定义则是在该点周围8个方向为‘x’。较为简单的dfs题....

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

思路：dfs，不过这次不是求有多少‘x’与之相连，求周长。先假设某个‘x’上下左右都没有‘x’，它的周长为4，然后四个方向上找到一个‘x’，周长--。（可能没说清楚，不过就是这么回事）

代码：

#include <iostream>
#include <algorithm>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <string>
using namespace std;
int visit[25][25];
char map[25][25];
int n,m,sum=0;
int d[8][2]= {{1,0},{1,-1},{0,-1},{-1,-1},
    {-1,0},{-1,1},{0,1},{1,1}};//搜索时的8个方向

bool inside(int x,int y)//搜索时判断是否在图内
{
    if(x<0||x>=n||y<0||y>=m)
        return false;
    return true;
}
void dfs(int x,int y)
{
    if(visit[x][y]==0&&map[x][y]=='X'&&inside(x,y))
    {
        visit[x][y]=1;
        int p=4;//默认该点周长为4
        if(map[x-1][y]=='X') --p;
        if(map[x][y-1]=='X') --p;
        if(map[x+1][y]=='X') --p;
        if(map[x][y+1]=='X') --p;
        sum+=p;
        for(int i=0; i<8; i++)//8个方向深搜
        {
            dfs(x+d[i][0],y+d[i][1]);
        }
    }
}
int main()
{
    int i,j,x,y;
    while(scanf("%d%d%d%d",&n,&m,&x,&y))//x,y为指定点的坐标
    {
        if(x+y+m+n==0)
            break;
        memset(visit,0,sizeof(visit));
        memset(map,0,sizeof(map));
        for(i=0; i<n; i++)
            for(j=0; j<m; j++)
            {
                cin>>map[i][j];
            }
        sum=0;
        dfs(x-1,y-1);//我是从map[0][0]开始记录的
        printf("%d\n",sum);
    }
    return 0;
}