BFS——Weed

Weed

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 131072/65536K (Java/Other)

Total Submission(s) : 15   Accepted Submission(s) : 3

Problem Description

Andrew has visited his garden for the last time many years ago. Today's property taxes are so high, so Andrew decided to sell his garden. The land was not cultivated for a long time and now it is probably a lot of weed on it. Andrew wants to remove everything from the ground before selling. Now he wants to estimate the amount of work.

The garden has the rectangular form and is divided into equal squares. Andrew's memory is phenomenal. He remembers which squares were occupied by the weed. For the purpose of simplicity, Andrew thinks that each square is either fully occupied by the weed or completely free from it. Andrew likes botany and he knows that if some square is free from the weed but at least two of its adjacent squares are occupied by the weed (two squares are adjacent if they have common side), that square will be also occupied by the weed soon. Andrew is pretty sure that during last years weed occupied every square possible. Please help Andrew to estimate how many squares is occupied by the weed.

Input

The first line of the input contains integers N and M (1 ≤ N, M ≤ 1000). Next N lines contain M characters each. Character X

denotes that the corresponding square is occupied by the weed. A period character (

.

) denotes an empty square.

Output

Print one integer denoting the number of squares occupied by the weed after so many years.

Example(s)

sample input	sample output
3 3 X.. .X. .X.	6

sample input	sample output
3 4 X..X .X.. .X..	12

题意：有一个只有杂草和空地的庄园。若杂草的上下左右四个地方有超过两个地方为杂草，则该空地不久之后也会变成杂草。求庄园内杂草的数量。

分析：用队列来放杂草，二维数组a记录该地是否为杂草，a[i][j]==0表示为杂草，空地a[x][y]初始化为-2。每一个杂草会使得它四方的a数组加1，当加到0时，空地变杂草并入队列。

AC代码

#include <cstdio>
#include <queue>
using namespace std;
int n,m;
char map[1005][1005];
int a[1005][1005];
int dir[4][2]={0,1,0,-1,-1,0,1,0};
struct node
{
    int x,y;
};
int ans;
queue<node> que;
int valid(int x,int y)
{
    if(0<=x&&x<n&&0<=y&&y<m)
        return 1;
    return 0;
}
void bfs()
{
    while(!que.empty())
    {
        node tmp=que.front();
        que.pop();
        for(int i=0;i<4;i++)
        {
            int fx=tmp.x+dir[i][0];
            int fy=tmp.y+dir[i][1];
            if(!valid(fx,fy))
                continue;
            if(a[fx][fy]==0)
                continue;
            a[fx][fy]++;
            if(a[fx][fy]==0)
            {
                ans++;
                que.push(node{fx,fy});
            }
        }
    }
    printf("%d
",ans);
}
int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        ans=0;
        while(!que.empty())
            que.pop();
        for(int i=0;i<n;i++)
            scanf("%s",map[i]);
        for(int i=0;i<n;i++)
            for(int j=0;j<m;j++)
        {
            if(map[i][j]=='X')
            {
                 a[i][j]=0;
                 ans++;
                 que.push(node{i,j});
            }
            else
                a[i][j]=-2;

        }
        bfs();
    }
    return 0;
}