BFS解决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
#include "iostream"
#include "string"
using namespace std;
int used[105][105];
char map[105][105];
int direction[8][2]={{-1,-1},{-1,0},{-1,1},{0,-1},{0,1},{1,-1},{1,0},{1,1}};
struct point
{
int x;
int y;
}queue[10005];
int si, sj, iColumn, iRow, count;
void bfs()
{
int head = 0, tail = 0;
used[si][sj] = 1;
queue[head].x = si; //搜索的初始x坐标
queue[head].y = sj; //搜索的初始y坐标
head++;
while(tail<head)
{
point temp1 = queue[tail];
tail++;
int k, m;
for(k=0; k<8; k++)
{
point temp2;
temp2.x = temp1.x+direction[k][0];
temp2.y = temp1.y+direction[k][1];
if(temp2.x>=0 && temp2.x<iRow && temp2.y>=0 && temp2.y<iColumn && !used[temp2.x][temp2.y] && map[temp2.x][temp2.y]=='W')
{
used[temp2.x][temp2.y] = 1;
queue[head] = temp2;
head++;
}
}
}
}
int main()
{
int i, j;
while(cin>>iRow>>iColumn)
{
memset(used, 0, sizeof(used));
count=0;
for(i=0; i<iRow; i++)
for(j=0; j<iColumn; j++)
cin>>map[i][j];
for(i=0; i<iRow; i++)
for(j=0; j<iColumn; j++)
{
if(map[i][j]=='W' && used[i][j]==0) //一旦遇到了W,并且没有被访问过就bfs(),count++;
{
si = i;
sj = j;
bfs();
count++;
}
}
cout<<count<<endl;
}
return 0;
}