c – 解决洪水般的难题的最小点击次数

我有网格N×M,其中每个单元格都用一种颜色着色.

当玩家点击颜色α的网格的任何单元格时,网格的顶部左上角的单元格β,接收颜色α,但不仅如此：所有这些通过连接到源的单元格仅使用颜色α或β的路径也接收颜色α.

单元格之间的连接应仅在水平和垂直方向上考虑以形成路径.例如,当玩家点击左侧图中突出显示的单元格时,网格会接收图形向右的着色.游戏的目标是使网格单色.

输入说明

The first line of the input consists of 2 integers N and M (1 ≤ N ≤ 4,1 ≤ M ≤ 5),which represent respectively the number of lines and the number of columns of the grid. The N lines following describe the initial configuration of the grid,representing each colour by an integer between 0 and 9. The input does not consist of any other line.

输出说明

Print a line containing a single integer that represents the minimum number of clicks that the player must do in order to make the grid monochromatic.

输入样品

4 5
01234
34567
67890
90123

2：

4 5
01234
12345
23456
34567

3：

4 5
00162
30295
45033
01837

输出样本

12

2：

7

3：

10

我试图找到一个解决方案与回溯(由于时间限制8秒和小尺寸的网格).但是超出了时间限制.有些人只是在0秒钟内完成.

还有一些其他算法来解决这个问题？

#include <stdio.h>
#include <string.h>

#define MAX 5
#define INF 999999999

typedef int signed_integer;

signed_integer n,m,mink;
bool vst[MAX][MAX];

signed_integer flood_path[4][2] = {
    {-1,0},{1,{0,1},-1}
};

//flood and paint all possible cells... the root is (i,j)
signed_integer flood_and_paint(signed_integer cur_grid[MAX][MAX],signed_integer i,signed_integer j,signed_integer beta,signed_integer alpha,signed_integer colors[]){
    //invalid cell
    if (vst[i][j] || i < 0 || i >= n || j < 0 || j >= m)
        return 0;

    //mark existent colors
    colors[cur_grid[i][j]] = 1;

    //only alpha and beta colors counts
    if (cur_grid[i][j] != beta && cur_grid[i][j] != alpha)
        return 0;

    //mark (i,j) as visited and change its color
    vst[i][j] = true;
    cur_grid[i][j] = alpha;

    //floodit !
    signed_integer ret = 1;
    for (signed_integer k = 0; k < 4; k++)
        ret += flood_and_paint(cur_grid,i + flood_path[k][0],j + flood_path[k][1],beta,alpha,colors);

    //how many cells change
    return ret;
}

void backtrack(signed_integer cur_grid[MAX][MAX],signed_integer k,signed_integer _cont,signed_integer alpha) {
    //bigger number of clicks for this solution ? ... getting back
    if(k >= mink)
        return;

    signed_integer colors[10];
    memset(vst,false,sizeof(vst));
    memset(colors,sizeof(colors));

    signed_integer beta = cur_grid[0][0];
    signed_integer cont = flood_and_paint(cur_grid,colors);

    //there are alpha colors to change and no beta colors to change
    colors[alpha] = 1;
    colors[beta]  = 0;

    //all squares on same color
    if (cont == n * m) {
        mink = k;
        return;
    }

    //this solution is equals to another ? ... getting back
    if (cont == _cont)
        return;

    ++k;//new click

    //copy this matrix and backtrack
    signed_integer copy[MAX][MAX];
    for (signed_integer c = 0; c < 10; ++c){
        if (colors[c] && c != cur_grid[0][0]) {
            memcpy(copy,cur_grid,n*m*sizeof(signed_integer));
            backtrack(copy,k,cont,c);
        }
    }
}

void cleanBuffer(){
     while (getchar() != '\n');
}

int main(void) {
    signed_integer grid[MAX][MAX];
    scanf("%d %d",&n,&m);
    for (signed_integer i = 0; i < n; ++i) {
        cleanBuffer();
        for (signed_integer j = 0; j < m; ++j){
            grid[i][j] = getchar() - '0';
        }
    }
    mink = INF;
    backtrack(grid,grid[0][0]);
    printf("%d\n",mink);
    return 0;
}

解决方法

高水平的改善

请注意,单元格是其原始颜色或最后选定的颜色.

这意味着我们可以用20位表示电路板的当前状态(4 * 5单元中的每个单元的标记是否包含原始颜色)以及范围为0到9的数字,给出最后选择的颜色.

这导致最多可以探索1000万个州.回溯功能可以避免如果达到已经访问的状态,则必须进行递归.我希望这个更改能够使您的解决方案更快.

水平提高

通过20bit掩码表示状态,最后一种颜色也使得状态更快更新,因为只需要更改2个数字,而不是全板的memcpy.

c – 解决洪水般的难题的最小点击次数

解决方法

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