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