【29.70%】【codeforces 723D】Lakes in Berland

time limit per test2 seconds

memory limit per test256 megabytes

inputstandard input

outputstandard 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 n, m 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.

【题解】



题意:和边相连的连通块不算lake,一开始有s个lake(s>=k),求最少要去掉多少个cell of lake才能让总的lake个数变为k;

先预处理将那些边上的连通块给置为1表示置为平地。但是要记录一下这些点的坐标。最后输出的时候还是要变回来。

然后i=1->n;j=1->m;寻找连通块。并记录各个连通块的起始坐标。获取这个连通块的cell个数设为size；

以size为关键字升序排序；

看需要减少多少个湖。就从小到大累加相应个数的size;

根据记录的起始坐标再次进行bfs(需要在一个初始且去掉边上的连通块的a数组的copy数组cpa进行);

最后把一开始去掉的边上的连通块重新置为是cell of lake;

最后输出整个矩阵;

#include <cstdio>
#include <queue>
#include <vector>
#include <algorithm>

using namespace std;

const int MAXN = 55;
const int dx[5] = { 0,1,-1,0,0 };
const int dy[5] = { 0,0,0,1,-1 };

int n, m,k,total = 0;
bool a[MAXN][MAXN] = { 0 }, cpa[MAXN][MAXN] = { 0 };
char s[MAXN];
struct data2
{
    int x,y;
    int size;
};

queue < data2 > dl;
vector < pair<int, int> > rest;
data2 qidian[MAXN*MAXN];
int cnt = 0;

void bfs1(int a0, int b0)
{
    rest.push_back(make_pair(a0, b0));
    a[a0][b0] = false;
    data2 temp;
    temp.x = a0, temp.y = b0;
    dl.push(temp);
    while (!dl.empty())
    {
        data2 temp1;
        temp1 = dl.front();
        int a1 = temp1.x, b1 = temp1.y;
        dl.pop();
        for (int i = 1; i <= 4; i++)
        {
            int a2 = a1 + dx[i];
            int b2 = b1 + dy[i];
            if (a[a2][b2])
            {
                a[a2][b2] = false;
                rest.push_back(make_pair(a2, b2));
                data2 temp2;
                temp2.x = a2;
                temp2.y = b2;
                dl.push(temp2);
            }
        }
    }
}

int bfs2(int a0, int b0,bool a[MAXN][MAXN])
{
    int num = 1;
    a[a0][b0] = false;
    data2 temp;
    temp.x = a0, temp.y = b0;
    dl.push(temp);
    while (!dl.empty())
    {
        data2 temp1;
        temp1 = dl.front();
        int a1 = temp1.x, b1 = temp1.y;
        dl.pop();
        for (int i = 1; i <= 4; i++)
        {
            int a2 = a1 + dx[i];
            int b2 = b1 + dy[i];
            if (a[a2][b2])
            {
                a[a2][b2] = false;
                num++;
                data2 temp2;
                temp2.x = a2;
                temp2.y = b2;
                dl.push(temp2);
            }
        }
    }
    return num;
}

bool cmp(data2 a, data2 b)
{
    return a.size < b.size;
}

int main()
{
    //freopen("F:\\rush.txt", "r", stdin);
    scanf("%d%d%d", &n, &m, &k);
    for (int i = 1; i <= n; i++)
    {
        scanf("%s", s);
        for (int j = 1; j <= m; j++)
            if (s[j - 1] == '.')
                a[i][j] = 1;
            else
                a[i][j] = 0;
    }
    for (int i = 1; i <= m; i++)//去掉边上的连通块。并记录那些连通块的每个cell的坐标之后方便回溯
        if (a[1][i])
            bfs1(1, i);
    for (int i = 1; i <= m; i++)
        if (a[n][i])
            bfs1(n, i);
    for (int i = 1; i <= n; i++)
        if (a[i][1])
            bfs1(i, 1);
    for (int i = 1; i <= n; i++)
        if (a[i][m])
            bfs1(i, m);
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= m; j++)//copy一下a数组
            cpa[i][j] = a[i][j];//这个cpa数组用来最后输出答案。删除操作也在这上面进行
    for (int i = 1;i <= n;i++)
        for (int j = 1; j <= m; j++)
            if (a[i][j])
            {
                cnt++;
                qidian[cnt].x = i; qidian[cnt].y = j;//记录这个湖的起点坐标
                qidian[cnt].size = bfs2(i, j,a);
            }
    total = cnt - k;
    sort(qidian + 1, qidian + 1 + cnt, cmp);//以湖的大小为关键字升序排
    int i = 0,zs = 0;
    while (total > 0)//看需要去掉几个湖
    {
        i++;
        zs += qidian[i].size;
        bfs2(qidian[i].x, qidian[i].y, cpa);
        total--;
    }
    int len = rest.size();
    for (int i = 0; i <= len - 1; i++)//把之前置为平地的重新置为cell of lake
    {
        int x = rest[i].first, y = rest[i].second;
        cpa[x][y] = 1;
    }
    printf("%d\n", zs);
    for (int i = 1; i <= n; i++)
    {
        for (int j = 1; j <= m; j++)
            if (cpa[i][j] == 1)
                putchar('.');
            else
                putchar('*');
        printf("\n");
    }
    return 0;
}