# Poj 2386 Lake Counting

qingshui23 分享于 2015-08-17

Lake Counting Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 24999 Accepted: 12619 Description Due to recent rains, water has pooled in va

Lake Counting Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 20782   Accepted: 10473 Description Due to recent rains, water has pooled i

``````                                                       ***Lake Counting***

Time Limit: 1000MS  Memory Limit: 65536K
Total Submissions: 24201  Accepted: 12216

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
``````

dfs  ，注意从8个方面进行搜索：

``````#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

int n,m,ans;
char map[110][110];

bool judge(int i,int j)
{
if(i>=n || i<0)
return false;
if(j>=m || j<0)
return false;
if(map[i][j] == '.')
return false;
return true;
}

void dfs(int i,int j)
{
map[i][j] = '.';
if(judge(i+1, j))
dfs(i+1, j);
if(judge(i, j+1))
dfs(i, j+1);
if(judge(i-1, j))
dfs(i-1, j);
if(judge(i, j-1))
dfs(i, j-1);
if(judge(i+1, j+1))
dfs(i+1, j+1);
if(judge(i-1, j-1))
dfs(i-1, j-1);
if(judge(i+1, j-1))
dfs(i+1, j-1);
if(judge(i-1, j+1))
dfs(i-1, j+1);
}

int main()
{
while(cin>>n>>m)
{
getchar();
ans=0;
for(int i=0; i<n; i++)
gets(map[i]);
for(int i=0; i<n; i++)
{
for(int j=0; j<m; j++)
{
if(map[i][j] == 'W')
{
ans++;
dfs(i,j);
}
}
}
cout<<ans<<endl;
}
return 0;
}
``````

click here~~ ***Lake Counting***Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 24201 Accepted: 12216 Desc

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