| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 22424 | Accepted: 11300 |
Description
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.
Input
* 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.
Output
* Line 1: The number of ponds in Farmer John‘s field.
Sample Input
10 12 W........WW. .WWW.....WWW ....WW...WW. .........WW. .........W.. ..W......W.. .W.W.....WW. W.W.W.....W. .W.W......W. ..W.......W.
Sample Output
3
Hint
OUTPUT DETAILS:
There are three ponds: one in the upper left, one in the lower left,and one along the right side.
Source
解题分析：
首先这道题目设计几个问题：
第一个就是怎么可以确定几个w是属于一块的。
第二个怎么搜索。
第三个怎么确保不会搜索重复的。
那么第一个和第二个是一个问题，就是深度优先搜索，然后，利用递归的方法，每次对于一个点是否有8个方向，如果有的话，继续8个方向的搜索，直到到达边界为止。
还有怎么确定不重复，那么就是我把每次找到的w都变味‘ . ‘ 就可以了。
具体看代码。
代码：
#include <iostream>
#include <cstdio>
#define MAXN 102
using namespace std;
char map[MAXN][MAXN];
int N, M;
int DG(int i, int j) {
if(i < 0 || i >= N || j < 0 || j >= M)
return 0;
if(map[i][j] == ‘.‘)
return 0;
map[i][j] = ‘.‘;
return 1 + DG(i-1, j-1) + DG(i-1, j) + DG(i-1, j+1)
+ DG(i, j-1) + DG(i, j+1) + DG(i+1, j-1) +
DG(i+1, j) + DG(i+1, j+1);
}
int main()
{
cin >> n >> m;
char flag;
for(int i = 0; i < n; i++) {
for(int j = 0; j < m; j++) {
cin >> flag;
map[i][j] = flag;
}
}
int count = 0;
for(int i = 0; i < n; i++) {
for(int j = 0; j < m; j++) {
if(DG(i, j) != 0)
count++;
}
}
cout << count << endl;
return 0;
}