POJ 2386 Lake Counting

Posted 2022-06-27 itdef

地址 http://poj.org/problem?id=2386

《挑战程序设计竞赛》习题

题目描述
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.

样例

 

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

算法1
将相同的水坑算在一起 并查集

C++ 代码

#include <iostream>
#include <set>

using namespace std;

#define MAX_NUM 110

int N, M;
char field[MAX_NUM+10][MAX_NUM + 10];
int fa[MAX_NUM*MAX_NUM];

//char field[10][12] = 
//  ‘W‘,‘.‘,‘.‘,‘.‘,‘.‘,‘.‘,‘.‘,‘.‘,‘.‘,‘W‘,‘W‘,‘.‘,
//  ‘.‘,‘W‘,‘W‘,‘W‘,‘.‘,‘.‘,‘.‘,‘.‘,‘.‘,‘W‘,‘W‘,‘W‘,
//  ‘.‘,‘.‘,‘.‘,‘.‘,‘W‘,‘W‘,‘.‘,‘.‘,‘.‘,‘W‘,‘W‘,‘.‘,
//  ‘.‘,‘.‘,‘.‘,‘.‘,‘.‘,‘.‘,‘.‘,‘.‘,‘.‘,‘W‘,‘W‘,‘.‘,
//  ‘.‘,‘.‘,‘.‘,‘.‘,‘.‘,‘.‘,‘.‘,‘.‘,‘.‘,‘W‘,‘.‘,‘.‘,
//  ‘.‘,‘.‘,‘W‘,‘.‘,‘.‘,‘.‘,‘.‘,‘.‘,‘.‘,‘W‘,‘.‘,‘.‘,
//  ‘.‘,‘W‘,‘.‘,‘W‘,‘.‘,‘.‘,‘.‘,‘.‘,‘.‘,‘W‘,‘W‘,‘.‘,
//  ‘W‘,‘.‘,‘W‘,‘.‘,‘W‘,‘.‘,‘.‘,‘.‘,‘.‘,‘.‘,‘W‘,‘.‘,
//  ‘.‘,‘W‘,‘.‘,‘W‘,‘.‘,‘.‘,‘.‘,‘.‘,‘.‘,‘.‘,‘W‘,‘.‘,
//  ‘.‘,‘.‘,‘W‘,‘.‘,‘.‘,‘.‘,‘.‘,‘.‘,‘.‘,‘.‘,‘W‘,‘.‘
//;


//===============================================
// union find
void init(int n)

    for(int i=1;i<=n;i++)
        fa[i]=i;

int get(int x)

    return fa[x]==x?x:fa[x]=get(fa[x]);//路径压缩,防止链式结构

void merge(int x,int y)

    fa[get(x)]=get(y);

//===========================================================



void Check(int x,int y)

    //上
    int xcopy = x - 1;
    if (xcopy >= 0 && x < N) 
        for (int add = -1; add <= 1; add++) 
            int ycopy = y + add;
            if (ycopy >= 0 && ycopy < M && field[xcopy][ycopy] == ‘W‘) 
                int idx = x * M + y;
                int anotherIdx = xcopy * M + ycopy;
                merge(idx, anotherIdx);
            
        
    

    //中
    xcopy = x;
    if (xcopy >= 0 && x < N) 
        for (int add = -1; add <= 1; add++) 
            if (add == 0) continue;
            int ycopy = y + add;
            if (ycopy >= 0 && ycopy < M && field[xcopy][ycopy] == ‘W‘) 
                int idx = x * M + y;
                int anotherIdx = xcopy * M + ycopy;
                merge(idx, anotherIdx);
            
        
    


    //下
    xcopy = x + 1;
    if (xcopy >= 0 && x < N) 
        for (int add = -1; add <= 1; add++) 
            int ycopy = y + add;
            if (ycopy >= 0 && ycopy < M && field[xcopy][ycopy] == ‘W‘) 
                int idx = x * M + y;
                int anotherIdx = xcopy * M + ycopy;
                merge(idx, anotherIdx);
            
        
    



int main()

    cin >> N >> M;
    //N = 10; M = 12;

    init(MAX_NUM*MAX_NUM);

    for (int i = 0; i < N; i++) 
        for (int j = 0; j < M; j++) 
            cin >> field[i][j];
            if (field[i][j] == ‘W‘) 
                //检查上下左右八个方向是否有坑
                Check(i,j);
            
        
    
    set<int> s;

    for (int i = 0; i < N; i++) 
        for (int j = 0; j < M; j++) 
            if (field[i][j] == ‘W‘) 
                int idx = i * M + j;
                //cout << "fa["<<idx << "] = "<< fa[idx] << endl;
                s.insert(get(idx));
            
        
    

    cout << s.size() << endl;

    return 0;



作者：defddr
链接：https://www.acwing.com/solution/acwing/content/3674/
来源：AcWing
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算法2
DFS 遍历 将坐标连续的坑换成. 计数+1

todo

C++ 代码