D. Arthur and Walls
time limit per test
2 seconds
memory limit per test
512 megabytes
input
standard input
output
standard 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 rectanglen×mconsisting of squares of size1×1. 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 integersn,m(1≤n,m≤2000) denoting the size of the Arthur apartments.
Followingnlines each containmsymbols — 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
Outputnrows each consisting ofmsymbols 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
………………..
题意：给出一个n*m的地图，由‘*’和‘.’号组成，，现在要将一些’.’改成’*’号使得所有局部的’.’号都能组成一个矩形，要保证修改的次数最少，最后输出改变后的矩形。
思路：最开始的思路是搜联通块，将联通块里面的’*’全部改成‘.’，但是题目范围较大，结果超时了。然后看到别人的是找一个基本元素块，n*m的矩形由这些元素块组成。发现：如果在一个2*2的方格内只有一个是‘*’那么就必须要将这个‘*’改成‘.’，这样bfs搜一遍即可。
代码：
#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 pfprintf#define DBGpf("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;elsea[i][j]=1;}}bfs();FRL(i,0,n){FRL(j,0,m){if (a[i][j])pf(".");elsepf("*");}pf("\n");}}return 0;}/*5 5*******.***.*.**…*******/
但是至少可以为自己的荷包省钱可以支些招，这点还是很现实的。
