标签：man sam put 次数 div void namespace and %s
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.
分析：注意水潭八个方向都是连通的，在主函数中对每个点进行dfs遍历，使用dfs遍历访问后做标记之后不再访问，主函数中dfs调用的次数即水池的个数
代码：
#include<iostream>
#include<cstdio>
using namespace std;
const int N = 105;
char mp[N][N];
int n, m;
void dfs(int x, int y) {
mp[x][y] = ‘.‘;
for(int i = -1; i <= 1 ; i++) {
for(int j = -1; j <= 1; j++) {
int nx = x + i, ny = y + j;
if(nx >= 0 && nx < n && ny >= 0 && ny < m && mp[nx][ny] == ‘W‘) {
dfs(nx, ny);
}
}
}
}
int main() {
scanf("%d%d", &n, &m);
for(int i = 0; i < n; i++) scanf("%s", mp[i]);
int count = 0;
for(int i = 0; i < n; i++) {
for(int j = 0; j < m; j++) {
if(mp[i][j] == ‘W‘) {
dfs(i, j);
count++;
}
}
}
printf("%d\n", count);
return 0;
}
POJ 2386 Lake Counting
标签：man sam put 次数 div void namespace and %s
原文地址：https://www.cnblogs.com/kindleheart/p/9296865.html
