洛谷-P1596 [USACO10OCT]Lake Counting S

洛谷-P1596 [USACO10OCT]Lake Counting S

原题链接:https://www.luogu.com.cn/problem/P1596


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;
}
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