Weed（BFS）

I - Weed

Time Limit:1000MS    Memory Limit:65536KB    64bit IO Format:%I64d & %I64u

SubmitStatusPracticeSGU 344

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 N x M 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.

Sample Input

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

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

对未长草的方格用深搜超时，对已长草的方格用广搜AC

AC CODE：

//Memory: 6135 KB         Time: 218 MS
//Language: GNU CPP (MinGW, GCC 4)         Result: Accepted

#include <iostream>
#include <string>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <queue>
#define LL long long
#define MAXI 2147483647
#define MAXL 9223372036854775807
#define eps (1e-8)
#define dg(i) cout << "*" << i << endl;
using namespace std;

struct Node
{
    int x, y;
};
int n, m, ans;
int a[4][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} }; //上下左右
int map[1001][1001];
//bool vis[1001][1001];
queue<Node> que;

bool in(int i, int j)
{
    return i > -1 && i < n && j < m && j > -1;
}

void BFS()
{
    while(!que.empty())
    {
        struct Node s = que.front();
        que.pop();
        for(int k = 0; k < 4; k++)
        {
            int ii = s.y + a[k][0];
            int jj = s.x + a[k][1];//cout << s.y <<" "<<s.x<<endl;
            if(in(ii, jj) && map[ii][jj] < 0)
            {
                map[ii][jj]++;
                if(!map[ii][jj])
                {//dg(1)
                    ans++;
                    struct Node tmp;
                    tmp.y = ii;
                    tmp.x = jj;
                    que.push(tmp);
                }
            }
        }
    }
}

int main()
{
    char c;
    int i, j;
    while(scanf("%d %d", &n, &m) != EOF)
    {
        struct Node s;
        ans = 0;
        memset(map, 0, sizeof(map));
        //memset(vis, 0, sizeof(vis));
        for(i = 0; i < n; i++)
        {
            scanf("%c", &c); //吃掉空格
            for(j = 0; j < m; j++)
            {
                scanf("%c", &c);
                if(c == 'X')
                {
                    map[i][j] = 0;
                    ans++;
                    s.y = i;
                    s.x = j;
                    que.push(s);
                }
                else map[i][j] = -2;
            }
        }
        BFS();
        printf("%d\n", ans);
    }
    return 0;
}