D. Arthur and Walls (CF 525 D 搜尋bfs)，arthurbfs

D. Arthur and Walls (CF 525 D 搜尋bfs)，arthurbfs

D. Arthur and Wallstime limit per test2 secondsmemory limit per test512 megabytesinputstandard inputoutputstandard output

Finally it is a day when Arthur has enough money for buying an apartment. He found a great option close to the center of the city with a nice price.

Plan of the apartment found by Arthur looks like a rectangle n × m consisting of squares of size 1 × 1. Each of those squares contains either a wall (such square is denoted by a symbol "*" on the plan) or a free space (such square is denoted on the plan by a symbol ".").

Room in an apartment is a maximal connected area consisting of free squares. Squares are considered adjacent if they share a common side.

The old Arthur dream is to live in an apartment where all rooms are rectangles. He asks you to calculate minimum number of walls you need to remove in order to achieve this goal. After removing a wall from a square it becomes a free square. While removing the walls it is possible that some rooms unite into a single one.

Input

The first line of the input contains two integers n, m (1 ≤ n, m ≤ 2000) denoting the size of the Arthur apartments.

Following n lines each contain m symbols — the plan of the apartment.

If the cell is denoted by a symbol "*" then it contains a wall.

If the cell is denoted by a symbol "." then it this cell is free from walls and also this cell is contained in some of the rooms.

Output

Output n rows each consisting of m symbols that show how the Arthur apartment plan should look like after deleting the minimum number of walls in order to make each room (maximum connected area free from walls) be a rectangle.

If there are several possible answers, output any of them.

Sample test(s)input

5 5.*.*.*****.*.*.*****.*.*.

output

.*.*.*****.*.*.*****.*.*.

input

6 7***.*.*..*.*.**.*.*.**.*.*.*..*...********

output

***...*..*...*..*...*..*...*..*...********

input

4 5............***..*..

output

....................

題意：給出一個n*m的地圖，由‘*’和‘.’號組成，現在要將一些'.'改成'*'號使得所有局部的'.'號都能組成一個矩形，要保證修改的次數最少，最後輸出改變後的矩形。

思路：最開始的思路是搜聯通塊，將聯通塊裡面的'*'全部改成‘.’，但是題目範圍較大，結果逾時了。然後看到別人的是找一個基本元素塊，n*m的矩形由這些元素塊組成。發現：如果在一個2*2的方格內只有一個是‘*’那麼就必須要將這個‘*’改成‘.’，這樣bfs搜一遍即可。

代碼：

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#include <string>#include <map>#include <stack>#include <vector>#include <set>#include <queue>#pragma comment (linker,"/STACK:102400000,102400000")#define maxn 2005#define MAXN 2005#define mod 1000000009#define INF 0x3f3f3f3f#define pi acos(-1.0)#define eps 1e-6#define lson rt<<1,l,mid#define rson rt<<1|1,mid+1,r#define FRE(i,a,b)  for(i = a; i <= b; i++)#define FREE(i,a,b) for(i = a; i >= b; i--)#define FRL(i,a,b)  for(i = a; i < b; i++)#define FRLL(i,a,b) for(i = a; i > b; i--)#define mem(t, v)   memset ((t) , v, sizeof(t))#define sf(n)       scanf("%d", &n)#define sff(a,b)    scanf("%d %d", &a, &b)#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)#define pf          printf#define DBG         pf("Hi\n")using namespace std;typedef pair<int,int> pa;int a[maxn][maxn];char mp[maxn][maxn];int n,m;bool Isok(int x,int y){    if (x>=0&&x<n&&y>=0&&y<m)        return true;    return false;}bool num(int x,int y){    int s=a[x][y]+a[x+1][y]+a[x][y+1]+a[x+1][y+1];    if (s==3) return true;    return false;}bool change(int x,int y){    if (a[x][y]) return false;    if (num(x,y))        return true;    if (x-1>=0&&num(x-1,y))        return true;    if (y-1>=0&&num(x,y-1))        return true;    if (x-1>=0&&y-1>=0&&num(x-1,y-1))        return true;    return false;}void bfs(){    int i,j;    queue<pa>Q;    while (!Q.empty()) Q.pop();    pa st,now;    FRL(i,0,n)    {        FRL(j,0,m)        {            if (change(i,j))            {                a[i][j]=1;                Q.push(make_pair(i,j));            }        }    }    while (!Q.empty())    {        st=Q.front(); Q.pop();        FRE(i,st.first-1,st.first+1)        {            FRE(j,st.second-1,st.second+1)            {                if (Isok(i,j)&&change(i,j))                {                    a[i][j]=1;                    Q.push(make_pair(i,j));                }            }        }    }}int main(){    int i,j;    while (~sff(n,m))    {        FRL(i,0,n)            scanf("%s",mp[i]);        FRL(i,0,n)        {            FRL(j,0,m)            {                if (mp[i][j]=='*')                    a[i][j]=0;                else                    a[i][j]=1;            }        }        bfs();        FRL(i,0,n)        {            FRL(j,0,m)            {                if (a[i][j])                    pf(".");                else                    pf("*");            }            pf("\n");        }    }    return 0;}/*5 5*******.***.*.**...*******/