NUC1158 Lake Counting【DFS】

Lake Counting

时间限制: 1000ms 内存限制: 65536KB

通过次数: 1总提交次数: 1

问题描述

Due to recent rains, water has pooled in various places in Farmer John's field, which is represented by a rectangle of N x M (1 <= N <= 100; 1 <= M <= 100) squares. Each square contains either water ('W') or dry land ('.'). Farmer John would like to figure out how many ponds have formed in his field. A pond is a connected set of squares with water in them, where a square is considered adjacent to all eight of its neighbors.

Given a diagram of Farmer John's field, determine how many ponds he has.

输入描述

* Line 1: Two space-separated integers: N and M * Lines 2..N+1: M characters per line representing one row of Farmer John's field. Each character is either 'W' or '.'. The characters do not have spaces between them.

输出描述

* Line 1: The number of ponds in Farmer John's field.

样例输入

10 12
W........WW.
.WWW.....WWW
....WW...WW.
.........WW.
.........W..
..W......W..
.W.W.....WW.
W.W.W.....W.
.W.W......W.
..W.......W.

样例输出

3

来源

USACO 2004 November Gold

提示

OUTPUT DETAILS: There are three ponds: one in the upper left, one in the lower left,and one along the right side.

问题分析：（略）

这个问题和《POJ2386 Lake Counting【DFS】》是同一个问题，代码直接用就AC了。

程序说明：参见参考链接。

参考链接：POJ2386 Lake Counting【DFS】

题记：程序做多了，不定哪天遇见似曾相识的。

AC的C++程序如下：

/* POJ2386 Lake Counting */

#include <stdio.h>
#include <string.h>

#define DIRECTSIZE 8

struct direct {
    int drow;
    int dcol;
} direct[DIRECTSIZE] =
    {{0, -1}, {0, 1}, {-1, 0}, {1, 0}, {-1, -1}, {-1, 1}, {1, -1}, {1, 1}};

#define MAXN 100

char grid[MAXN+2][MAXN+2];

void dfs(int row, int col)
{
    int i;

    for(i=0; i<DIRECTSIZE; i++) {
        int nextrow = row + direct[i].drow;
        int nextcol = col + direct[i].dcol;

        if(grid[nextrow][nextcol] == 'W') {
            grid[nextrow][nextcol] = '.';

            dfs(nextrow, nextcol);
        }
    }
}

int main(void)
{
    int m, n, count, i, j;

    while(scanf("%d%d", &m, &n) != EOF) {
        // 清零：边界清零
        memset(grid, 0, sizeof(grid));

        // 读入数据
        for(i=1; i<=m; i++)
            scanf("%s", grid[i]+1);

        // 计数清零
        count = 0;

        // 深度优先搜索
        for(i=1; i<=m; i++)
            for(j=1; j<=n; j++)
                if(grid[i][j] == 'W') {
                    count++;
                    grid[i][j] = '.';
                    dfs(i, j);
                }

        // 输出结果
        printf("%d\n", count);
    }

    return 0;
}