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  • 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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  • 原文地址:https://www.cnblogs.com/yuzec/p/14166420.html
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