POJ2386 簡單的深搜
Lake Counting
|
Time Limit: 1000MS |
Memory Limit: 65536K |
|
Total Submissions: 20765 |
Accepted: 10463 |
Description
Due to recent rains, water has pooled in various placesin 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 pondshave formed in his field. A pond is a connected set of squares with water inthem, 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'sfield. Each character is either 'W' or '.'. The characters do not have spacesbetween 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
程式碼：
#include<stdio.h>
int N,M;
char field[110][110];
void dfs(int x, int y)
{
intdx,dy;
intnx,ny;
field[x][y]='.';
for(dx=-1;dx<=1;dx++)//
{
for(dy=-1;dy<=1;dy++)//向x方向移動dx,向y方向移動dy，移動後為nx，ny
{
nx= x + dx;
ny= y + dy;
if(nx>=0&& ny>=0 && nx<N && ny<M &&field[nx][ny]=='W')
//判斷移動後的點是不是在院子內，以及是否為W
{
dfs(nx,ny);
}
}
}
return;
}
int main()
{
inti,j;
intnum = 0;
scanf("%d",&N);
scanf("%d",&M);
for(i=0;i<N;i++)//N代表行
{
scanf("%s",&field[i]);
}
for(i=0;i<N;i++)
{
for(j=0;j<M;j++)
{
if(field[i][j]=='W')//從有W的地方開始進行DFS
{
dfs(i,j);
num++;
}
}
}
printf("%d\n",num);
return0 ;
}