标签：P1596 John .. int WW ..... USACO10OCT Farmer Counting
洛谷-P1596 [USACO10OCT]Lake Counting S
题目描述
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.
由于近期的降雨，雨水汇集在农民约翰的田地不同的地方。我们用一个NxM(1<=N<=100;1<=M<=100)网格图表示。每个网格中有水('W') 或是旱地('.')。一个网格与其周围的八个网格相连，而一组相连的网格视为一个水坑。约翰想弄清楚他的田地已经形成了多少水坑。给出约翰田地的示意图，确定当中有多少水坑。
输入格式
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.
第1行：两个空格隔开的整数：N 和 M 第2行到第N+1行：每行M个字符，每个字符是'W'或'.'，它们表示网格图中的一排。字符之间没有空格。
输出格式
Line 1: The number of ponds in Farmer John's field.
一行：水坑的数量
输入输出样例
输入 #1
10 12
W........WW.
.WWW.....WWW
....WW...WW.
.........WW.
.........W..
..W......W..
.W.W.....WW.
W.W.W.....W.
.W.W......W.
..W.......W.
输出 #1
3
说明/提示
OUTPUT DETAILS: There are three ponds: one in the upper left, one in the lower left, and one along the right side.
C++代码
#include <iostream>
using namespace std;
char a[105][105];
int n, m, ans, dir[][2]={{-1, -1}, {-1, 0}, {-1, 1},
{0, -1}, {0, 1}, {1, -1}, {1, 0}, {1, 1}};
void dfs(int x, int y) {
int dx, dy;
a[x][y] = '.';
for (int i=0; i<8; ++i) {
dx = x + dir[i][0];
dy = y + dir[i][1];
if (dx>=0&&dx<n&&dy>=0&&dy<m&&a[dx][dy]=='W')
dfs(dx, dy);
}
}
int main() {
cin >> n >> m;
for (int i=0; i<n; ++i)
for (int j=0; j<m; ++j)
cin >> a[i][j];
for (int i=0; i<n; ++i)
for (int j=0; j<m; ++j)
if (a[i][j] == 'W') {
dfs(i, j);
++ans;
}
cout << ans << endl;
return 0;
}
标签：P1596,John,..,int,WW,.....,USACO10OCT,Farmer,Counting 来源： https://www.cnblogs.com/yuzec/p/14166420.html
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