D. Arthur and Walls (CF 525 D search BFS)

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

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

Pla N of the apartment found by Arthur looks like a rectangle  ").

The apartment is a maximal connected area consisting of free squares. Squares is considered adjacent if they share a common side.

The old Arthur dream was to live in a apartment where all rooms is 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 was possible that some rooms unite to a single one.

Input

The first line of the input contains the 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's this cell is free from walls and also this cell is contained in so Me of the rooms.

Output

Output n rows each consisting of m symbols so show how the Arthur apartment plan should look like Afte R Deleting the minimum number of walls in order to make each guest (maximum connected area free from walls) is a rectangle.

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

Sample Test (s) input

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

Output

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

Input

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

Output

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

Input

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

Output

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

Test instructions: Give a map of n*m, by ' * ' and '. ' Number, and now we're going to have some '. ' Change the ' * ' to make all the local '. ' Can form a rectangle, to ensure the least number of changes, and finally output the changed rectangle.

Idea: The first idea is to search the Unicom block, the Unicom block inside the ' * ' all changed to '. ', but the scope of the topic is larger, the result has timed out. And then see someone else is looking for a block of basic elements, the n*m rectangle consists of these element blocks. Discovery: If only one in a 2*2 box is ' * ' then you have to change this ' * ' to '. ' So that the bfs can be searched again.

Code:

#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 RSO  N 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, si Zeof (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 PA ir<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*******.***.*.**...*******/

D. Arthur and Walls (CF 525 D search BFS)