Baby Ming and Matrix games

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 849    Accepted Submission(s): 211

Problem Description

These few days, Baby Ming is addicted to playing a matrix game.

Given a n∗m matrix, the character in the matrix(i∗2,j∗2) (i,j=0,1,2...) are the numbers between 0−9. There are an arithmetic sign (‘+’, ‘-‘, ‘∗’, ‘/’) between every two adjacent numbers, other places in the matrix fill with ‘#’.

The question is whether you can find an expressions from the matrix, in order to make the result of the expressions equal to the given integer sum. (Expressions are calculated according to the order from left to right)

Get expressions by the following way: select a number as a starting point, and then selecting an adjacent digital X to make the expressions, and then, selecting the location of X for the next starting point. (The number in same place can’t be used twice.)

Input

In the first line contains a single positive integer T, indicating number of test case.

In the second line there are two odd numbers n,m, and an integer sum(−1018<sum<1018, divisor 0 is not legitimate, division rules see example)

In the next n lines, each line input m characters, indicating the matrix. (The number of numbers in the matrix is less than 15)

1≤T≤1000

Output

Print Possible if it is possible to find such an expressions.

Print Impossible if it is impossible to find such an expressions.

Sample Input

3
3 3 24
1*1
+#*
2*8
1 1 1
1
3 3 3
1*0
/#*
2*6

Sample Output

Possible
Possible
Possible

Hint

The first sample：1+2*8=24
The third sample：1/2*6=3

题意：在这个矩阵内是否可以找到一个表达式的值等于sum;

思路：dfs找到所有的表达式，暴力一发；

ＡＣ代码：

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
using namespace std;
double sum;
int t,n,m,flag;
int vis[100][100],dir[4][2]={-2,0,2,0,0,2,0,-2};
double num[100][100];
char s[100][100];
int dfs(int x,int y,double ans)
{
    vis[x][y]=1;
    if(fabs(ans-sum)<=0.000000001)flag=1;
    for(int i=0;i<4;i++)
    {
       int fx=x+dir[i][0],fy=y+dir[i][1];
       int px=x+dir[i][0]/2,py=y+dir[i][1]/2;
       if(fx>=1&&fx<=n&&fy<=m&&fy>=1&&vis[fx][fy]==0&&s[fx][fy]!=‘#‘)
       {
           if(s[px][py]==‘+‘)
           dfs(fx,fy,ans+num[fx][fy]);
           else if(s[px][py]==‘*‘)
            dfs(fx,fy,ans*num[fx][fy]);
           else if(s[px][py]==‘-‘)
            dfs(fx,fy,ans-num[fx][fy]);
           else if(s[px][py]==‘/‘&&num[fx][fy]!=0)
            dfs(fx,fy,ans/num[fx][fy]);
       }
    }
    vis[x][y]=0;
    return 0;
}
int main()
{
    scanf("%d",&t);
    while(t--)
    {
        flag=0;
        memset(vis,0,sizeof(vis));
        memset(num,-1,sizeof(num));
        scanf("%d%d%lf",&n,&m,&sum);
        for(int i=1;i<=n;i++)
        {
            scanf("%s",s[i]+1);
        }
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=m;j++)
            {
                if(s[i][j]>=‘0‘&&s[i][j]<=‘9‘)
                {
                    num[i][j]=s[i][j]-‘0‘;
                }
            }
        }
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=m;j++)
            {
                if(s[i][j]>=‘0‘&&s[i][j]<=‘9‘)
                {
                    dfs(i,j,num[i][j]);
                }
            }
        }
        if(flag)printf("Possible\n");
        else printf("Impossible\n");
    }
    return 0;
}