POJ3051 Satellite Photographs【DFS】

Satellite Photographs
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 6207 Accepted: 2924

Description

Farmer John purchased satellite photos of W x H pixels of his farm (1 <= W <= 80, 1 <= H <= 1000) and wishes to determine the largest 'contiguous' (connected) pasture. Pastures are contiguous when any pair of pixels in a pasture can be connected by traversing adjacent vertical or horizontal pixels that are part of the pasture. (It is easy to create pastures with very strange shapes, even circles that surround other circles.)

Each photo has been digitally enhanced to show pasture area as an asterisk ('*') and non-pasture area as a period ('.'). Here is a 10 x 5 sample satellite photo:

.......**
...***
........
...
..*.

This photo shows three contiguous pastures of 4, 16, and 6 pixels. Help FJ find the largest contiguous pasture in each of his satellite photos.

Input

• Line 1: Two space-separated integers: W and H

• Lines 2..H+1: Each line contains W "*" or "." characters representing one raster line of a satellite photograph.

Output

• Line 1: The size of the largest contiguous field in the satellite photo.

Sample Input

10 5
.......**
...***
........
...
..*.

Sample Output

16

Source

USACO 2005 November Bronze

问题链接：POJ3051 Satellite Photographs
问题简述：（略）
问题分析：
    计算最大的联通区域。用递归的DFS函数实现，逻辑代码最为简洁。
程序说明：（略）
参考链接：（略）
题记：（略）

AC的C语言程序如下：

/* POJ3051 Satellite Photographs  */

#include <stdio.h>

#define N 1000
char s[N][N + 1];
int w, h;

int dfs(int row, int col)
{
    if(row < 0 || row >= h || col < 0 || col >= w || s[row][col] != '*')
        return 0;

    int ans = 1;
    s[row][col] = '.';
    ans += dfs(row + 1, col) + dfs(row - 1, col) + dfs(row, col + 1) + dfs(row, col - 1);

    return ans;
}

int main(void)
{
    int ans, maxans, i, j;
    while(~scanf("%d%d", &w, &h)) {
        for(i = 0; i < h; i++)
            scanf("%s", s[i]);

        maxans = 0;
        for(i = 0; i < h; i++)
            for(j = 0; j < w; j++)
                if(s[i][j] == '*') {
                    ans = dfs(i, j);
                    maxans = maxans < ans ? ans : maxans;
                }

        printf("%d\n", maxans);
    }

    return 0;
}