D. Lakes in Berland
time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

The map of Berland is a rectangle of the size n × m, which consists of cells of size 1 × 1. Each cell is either land or water. The map is surrounded by the ocean.

Lakes are the maximal regions of water cells, connected by sides, which are not connected with the ocean. Formally, lake is a set of water cells, such that it's possible to get from any cell of the set to any other without leaving the set and moving only to cells adjacent by the side, none of them is located on the border of the rectangle, and it's impossible to add one more water cell to the set such that it will be connected with any other cell.

You task is to fill up with the earth the minimum number of water cells so that there will be exactly k lakes in Berland. Note that the initial number of lakes on the map is not less than k.

Input

The first line of the input contains three integers nm and k (1 ≤ n, m ≤ 50, 0 ≤ k ≤ 50) — the sizes of the map and the number of lakes which should be left on the map.

The next n lines contain m characters each — the description of the map. Each of the characters is either '.' (it means that the corresponding cell is water) or '*' (it means that the corresponding cell is land).

It is guaranteed that the map contain at least k lakes.

Output

In the first line print the minimum number of cells which should be transformed from water to land.

In the next n lines print m symbols — the map after the changes. The format must strictly follow the format of the map in the input data (there is no need to print the size of the map). If there are several answers, print any of them.

It is guaranteed that the answer exists on the given data.

Examples
input
5 4 1
****
*..*
****
**.*
..**
output
1
****
*..*
****
****
..**
input
3 3 0
***
*.*
***
output
1
***
***
***
Note

In the first example there are only two lakes — the first consists of the cells (2, 2) and (2, 3), the second consists of the cell (4, 3). It is profitable to cover the second lake because it is smaller. Pay attention that the area of water in the lower left corner is not a lake because this area share a border with the ocean.

题目链接:D. Lakes in Berland

题意就是给一个n*m的地图,其中边缘地区邻接着海,不能算一个湖,里面的每一个连通块都算一个湖,求最少填掉多少块小格子使得湖的数量变成k(起始湖数量大于等于k)

没见过这么水的D题……先DFS求不邻接大海的连通块(这里如果用并查集求连通块就判断邻接大海就没有DFS这么简便),然后把每一块中DFS的起始点坐标和湖的面积组成结构体记录并按照湖的面积排序,再贪心地再写一个DFS填充面积小的湖,然而地没注释调试信息和调用的函数名字搞错WA了很多次

代码:

#include <stdio.h>
#include <bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define CLR(arr,val) memset(arr,val,sizeof(arr))
#define LC(x) (x<<1)
#define RC(x) ((x<<1)+1)
#define MID(x,y) ((x+y)>>1)
typedef pair<int,int> pii;
typedef long long LL;
const double PI=acos(-1.0);
const int N=55;
struct info
{
    int need;
    int x,y;
    bool operator<(const info &t)const
    {
        return need>t.need;
    }
};
priority_queue<info>Q;

char pos[N][N];
int vis[N][N];
int n,m;
int dir[4][2]={{1,0},{-1,0},{0,1},{0,-1}};
int is_lake,once;

inline bool check(int x,int y)
{
    return x>=0&&x<n&&y>=0&&y<m&&pos[x][y]=='.'&&!vis[x][y];
}
void dfs(int x,int y)
{
    vis[x][y]=1;
    ++once;
    if(x==0||y==0||x==n-1||y==m-1)
        is_lake=0;
    for (int i=0; i<4; ++i)
    {
        int xx=x+dir[i][0];
        int yy=y+dir[i][1];
        if(check(xx,yy))
            dfs(xx,yy);
    }
}
void kuosan(int x,int y)
{
    pos[x][y]='*';
    for (int i=0; i<4; ++i)
    {
        int xx=x+dir[i][0];
        int yy=y+dir[i][1];
        if(xx>=0&&xx<n&&yy>=0&&yy<m&&pos[xx][yy]=='.')
            kuosan(xx,yy);
    }
}
int main(void)
{
    int i,j,k;
    while (~scanf("%d%d%d",&n,&m,&k))
    {
        CLR(vis,0);
        for (i=0; i<n; ++i)
            scanf("%s",pos[i]);
        int sc=0;
        for (i=0; i<n; ++i)
        {
            for (j=0; j<m; ++j)
            {
                if(pos[i][j]=='.'&&!vis[i][j])
                {
                    is_lake=1;
                    once=0;
                    dfs(i,j);
                    if(is_lake)
                        Q.push((info){once,i,j}),++sc;
                }
            }
        }
        int r=0;
        info now;
        while (sc>k)
        {
            now=Q.top();
            Q.pop();
            kuosan(now.x,now.y);
            r+=now.need;
            --sc;
        }
        printf("%d\n",r);
        for (i=0; i<n; ++i)
            printf("%s\n",pos[i]);
    }
    return 0;
}

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