poj 2386 Lake Counting

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.

Source

USACO 2004 November

/*
Problem: 2386  User: saruka
Memory: 812K  Time: 16MS
Language: G++  Result: Accepted
*/
#include<cstdio>
#include<cstring>
using namespace std;
inline int getint()
{
int r = 0, k = 1;
char c = getchar();
for(; c < '0' || c > '9'; c = getchar())
if(c == '-') k = -1;
for(; c >= '0' && c <= '9'; c = getchar())
r = r * 10 + c - '0';
return r * k;
}
const int dx[8] = {-1, 1, 0, 0, -1, 1, -1, 1};
const int dy[8] = {0, 0, -1, 1, -1, -1, 1, 1};
const int maxn = 105;
int n, m, xx, yy, sum = 0;
char graph[maxn][maxn];
bool vis[maxn][maxn];
void init()
{
memset(graph, 0, sizeof(graph));
memset(vis, 0, sizeof(vis));
}
void DFS(int x, int y)
{
vis[x][y] = true;
for(int i = 0; i < 8; i++)
{
xx = x + dx[i];
yy = y + dy[i];
if((graph[xx][yy] == 'W') && (!vis[xx][yy]))
{
DFS(xx, yy);
}
}
}
int main()
{
init();
n = getint();
m = getint();
for(int i = 0; i < n; i++)
{
scanf("%s", graph[i]);
}
for(int i = 0; i < n; i++)
{
for(int j = 0; j < m; j++)
{
if((graph[i][j] == 'W') && (!vis[i][j]))
{
DFS(i, j);
sum++;
}
}
}
printf("%d\n", sum);
return 0;
}

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