Google Code Jam 2014-Qualification Round-Problem C. Minesweeper Master

Problem

Minesweeper is a computer game that became popular in the 1980s, and is still included in some versions of the Microsoft Windows operating system. This problem has a similar idea, but it does not assume you have played Minesweeper.

In this problem, you are playing a game on a grid of identical cells. The content of each cell is initially hidden. There are M mines hidden in M different cells of the grid. No other cells contain mines. You may click on any cell to reveal it. If the revealed cell contains a mine, then the game is over, and you lose. Otherwise, the revealed cell will contain a digit between 0 and 8, inclusive, which corresponds to the number of neighboring cells that contain mines. Two cells are neighbors if they share a corner or an edge. Additionally, if the revealed cell contains a 0, then all of the neighbors of the revealed cell are automatically revealed as well, recursively. When all the cells that don't contain mines have been revealed, the game ends, and you win.

For example, an initial configuration of the board may look like this ('*' denotes a mine, and 'c' is the first clicked cell):

*..*...**.....*.......c..*............*...........

There are no mines adjacent to the clicked cell, so when it is revealed, it becomes a 0, and its 8 adjacent cells are revealed as well. This process continues, resulting in the following board:

*..*...**.1112*.....00012*....00001111*.00000001..

At this point, there are still un-revealed cells that do not contain mines (denoted by '.' characters), so the player has to click again in order to continue the game.

You want to win the game as quickly as possible. There is nothing quicker than winning in one click. Given the size of the board (R x C) and the number of hidden mines M, is it possible (however unlikely) to win in one click? You may choose where you click. If it is possible, then print any valid mine configuration and the coordinates of your click, following the specifications in the Output section. Otherwise, print "Impossible".

Input

The first line of the input gives the number of test cases, T. T lines follow. Each line contains three space-separated integers: R, C, and M.

Output

For each test case, output a line containing "Case #x:", where x is the test case number (starting from 1). On the following R lines, output the board configuration with C characters per line, using '.' to represent an empty cell, '*' to represent a cell that contains a mine, and 'c' to represent the clicked cell.

If there is no possible configuration, then instead of the grid, output a line with"Impossible" instead. If there are multiple possible configurations, output any one of them.

Limits

0 ≤ M < R * C.

Small dataset

1 ≤ T ≤ 230.
1 ≤ R, C ≤ 5.

Large dataset

1 ≤ T ≤ 140.
1 ≤ R, C ≤ 50.

Sample

Input 
 
Output 
 

55 5 233 1 12 2 14 7 310 10 82

Case #1:ImpossibleCase #2:c.*Case #3:ImpossibleCase #4:......*.c....*.........*....Case #5:**********************************....*****.....*****.c...*****....*********************************

题意：扫雷游戏中，给出矩形大小R*C，以及雷数M，求是否存在一种摆法，使得一次点击就赢，若存在，输出任一种合理的摆法。（方格内数字表示周围8个格的雷数，若点开方格内数字为0，则向周围八个格递归扩展）

分析：考虑了很多种摆雷的策略，最后终于找到一种正确的。计算空格数R*C-M，先把起始点放在左上角，这样需要的周围三个空格即可递归扩展，具体方法如下：

1、从大到小枚举每一行的空格数，不大于上一行空格数，不能小于2,。（一行只有一个空格，这个空格无法被扩展到）

2、第二行空格数与第一行相同。（保证第一行最后一个空格不是凸出来的，否则无法被扩展到）

3、记录已经用了的空格数，若其等于空格总数，找到一个合法摆法。

需特殊处理的情况：

1、全部是雷，R*C=M

2、只有一行或一列，R=1或C=1

3、只有一个空格，R*C=M+1

代码：

#include<cstdio>#include<cstring>#include<cstdlib>#include<cmath>#include<cstring>#include<string>#include<iostream>#include<queue>#include<stack>using namespace std;const int N=60;int r,c,m,sp;int col[N];bool check(int row,int pre,int now){    if(sp==1)    {        col[0] = 1;        return true;    }    if(row>=r)    {        return now==sp;    }    if(now>sp) return false;    if(now==sp) return true;    if(row==1)    {        col[row] = pre;        return check(row+1,pre,now+pre);    }    for(int i=pre; i>=2; i--)    {        col[row] = i;        if(check(row+1,i,now+i))            return true;        col[row] = 0;    }    return false;}int main(){freopen("C-large.in","r",stdin);freopen("C-large.out","w",stdout);int t;scanf("%d",&t);for(int cnt=1;cnt<=t;cnt++){        scanf("%d%d%d",&r,&c,&m);    printf("Case #%d:\n",cnt);    if(m==r*c)        {            printf("Impossible\n");            continue;        }        if(r==1 || c==1)        {            int ls = r*c-m;            for(int i=0; i<r; i++)            {                for(int j=0; j<c; j++)                {                    if(ls>0)                    {                        if(i==0 && j==0) printf("c");                        else printf(".");                        ls--;                    }                    else printf("*");                }                printf("\n");            }            continue;        }        sp = r*c-m;        memset(col,0,sizeof(col));        if(check(0,min(sp/2,c),0))        {            for(int i=0; i<r; i++)            {                for(int j=0; j<c; j++)                {                    if(i==0 && j==0) printf("c");                    else if(j<col[i]) printf(".");                    else printf("*");                }                printf("\n");            }        }        else            printf("Impossible\n");} }