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
View Code
#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; }
posted on 2011-09-23 22:37 More study needed. 阅读(199) 评论(0) 编辑 收藏 举报