Lake Counting
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 49414 | Accepted: 24273 |
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.
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
Hint
OUTPUT DETAILS:
There are three ponds: one in the upper left, one in the lower left,and one along the right side.
#include <iostream>
#include <algorithm>
#include <cmath>
using namespace std;
char a[][];
int n,m;
void dfs(int x,int y)
{
a[x][y]='.';
for(int dx=-;dx<=;dx++){
for(int dy=-;dy<=;dy++){
int nx=x+dx,ny=y+dy;
if(nx>=&&nx<n&&ny>=&&ny<m&&a[nx][ny]=='W'){
dfs(nx,ny);
}
}
}
return ;
}
void solve()
{
int res=;
for(int i=;i<n;i++){
for(int j=;j<m;j++){
if(a[i][j]=='W'){
dfs(i,j);
res++;
}
}
}
cout<<res<<endl;
}
int main()
{
while(cin>>n>>m){
for(int i=;i<n;i++){
for(int j=;j<m;j++){
cin>>a[i][j];
}
}
solve();
}
return ;
}
自己再熟悉一遍：
#include <iostream>
#include <cmath>
#include <algorithm>
#include <string>
#include <cstring>
using namespace std;
int n,m;
char a[][];
int res;
int nx,ny;
void dfs(int x,int y)
{
//八个方向搜索
for(int i=-;i<=;i++){
for(int j=-;j<=;j++){
nx=x+i,ny=y+j;
if(nx>=&&nx<n&&ny>=&&ny<m&&a[nx][ny]=='W'){
a[nx][ny]='.';
dfs(nx,ny);
}
}
}
}
void solve()
{
res=;
//每次从发现W的时候开始搜索由于一簇只能算一个地方所以res计数只加一次
for(int i=;i<n;i++){
for(int j=;j<m;j++){
if(a[i][j]=='W'){
res++;
dfs(i,j);
}
}
}
}
int main()
{
while(cin>>n>>m){
memset(a,,sizeof(a));
for(int i=;i<n;i++){
for(int j=;j<m;j++){
cin>>a[i][j];
}
}
solve();
cout<<res<<endl;
}
return ;
}