DFS解決POJ 2386
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.
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.
* 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
View Code
#include <stdio.h>
char M[101][101];
int Used[101][101];
int direction[8][2]={{-1,-1},{-1,0},{-1,1},{0,-1},{0,1},{1,-1},{1,0},{1,1}};
int i,j,p;
int col,row;
int iCount=0;
void DFS(int x,int y)
{
int k;
for(k=0;k<8;k++)
{
int tx= x+direction[k][0];
int ty= y+direction[k][1];
if(tx >= 0 && tx < row && ty >= 0 && ty < col && M[tx][ty] == 'W' && Used[tx][ty]==0)
{
Used[tx][ty]=1;
DFS(tx,ty);
}
}
}
int main()
{
scanf("%d%d",&row,&col);
for(i=0;i<row;i++)
scanf("%s",M[i]);
for(i=0;i<row;i++)
{
for(j=0;j<col;j++)
{
if(M[i][j]=='W' && Used[i][j] == 0)
{
Used[i][j]=1; //將其本身賦為已訪問過(注意:這一點容易被忽略)
DFS(i,j);
iCount++;
}
}
}
printf("%d\n",iCount);
}
posted on 2011-09-23 22:35 More study needed. 閱讀(154) 評論(0) 收藏 舉報
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