poj 2386 Lake Counting.md

题目

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 Given a diagram of Farmer John's field, determine how many ponds he has.

输入描述

*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.

输出描述

*Line 1: The number of ponds in Farmer John's field.

输入例子

10 12
W........WW.
.WWW.....WWW
....WW...WW.
.........WW.
.........W..
..W......W..
.W.W.....WW.
W.W.W.....W.
.W.W......W.
..W.......W.

输出例子

• 3

Hint

OUTPUT DETAILS:

There are three ponds: one in the upper left, one in the lower left,and one along the right side.

参考答案

#include<iostream>
#include<queue>
using namespace std;

struct Node
{
  int x;
  int y;
};

int dir[8][2]={{0,1},{0,-1},{1,0},{-1,0},{1,-1},{-1,1},{1,1},{-1,-1}};
char map[105][105];
int num;
int n,m;

void BFS(int x,int y)
{
  queue<Node>q;
  int tx,ty;
  int i,j;
  Node t,p;
  t.x=x;
  t.y=y;
  q.push(t);
  while(!q.empty())
  {
    p=q.front();
    q.pop();
    for(i=0;i<8;i++)
    {
      tx=p.x+dir[i][0];
      ty=p.y+dir[i][1];
      if(tx>=0&&tx<n&&ty>=0&&ty<m&&map[tx][ty]=='W')
      {
        t.x=tx;
        t.y=ty;
        q.push(t);
        map[tx][ty]='.';
      }
    }
  }
}

int main()
{
  int i,j;
  while(scanf("%d%d",&n,&m)!=EOF)
  {
    num=0;
    for(i=0;i<n;i++)
    {
      scanf("%s",map[i]);
    }
    for(i=0;i<n;i++)
    {
      for(j=0;j<m;j++)
      {
        if(map[i][j]=='W')
        {
          num++;
          BFS(i,j);
        }
      }
    }
    printf("%d\n",num);
  }
  return 0;
}