Baby Ming and Matrix games

Baby Ming and Matrix games

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1687    Accepted Submission(s): 481

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

 

#include <bits/stdc++.h>

using namespace std;
typedef long long ll;
const int maxn=1e3;
char str[maxn][maxn];
int T,n,m;
ll sum;
bool flag=false;
int dx[5]={0,2,-2,0,0},dy[5]={0,0,0,2,-2};
bool ok(char ch){
     if(ch=='+'||ch=='-'||ch=='*'||ch=='/')return true;
     else return false;
}
double cal(double a,double b,char c){
     switch(c){
        case '+':return a+b;
        case '-':return a-b;
        case '*':return a*b;
        case '/':return a/b;
     }
}
void dfs(int x,int y,double tot){
      if(flag)return;
      if(fabs(tot-sum)<=1e-6){
            flag=true;
            return;
      }
      for(int i=1;i<=4;i++){
            int cx=x+dx[i],cy=y+dy[i];
            if(cx<1||cy<1||cx>n||cy>m)continue;
            if(!isdigit(str[cx][cy]))continue;
            if(!ok(str[x+dx[i]/2][y+dy[i]/2]))continue;
            if(str[x+dx[i]/2][y+dy[i]/2]=='/'&&str[cx][cy]=='0')continue;
            char q=str[cx][cy];
            double ans=cal(tot,str[cx][cy]-'0',str[x+dx[i]/2][y+dy[i]/2]);
            str[cx][cy]='#';
            dfs(cx,cy,ans);
            str[cx][cy]=q;
      }
}
int main()
{
    cin>>T;
    while(T--){
        flag=false;
        cin>>n>>m>>sum;
        for(int i=1;i<=n;i++)scanf("%s",str[i]+1);
        /*for(int i=1;i<=n;i++){
            for(int j=1;j<=m;j++){
                printf("%c",str[i][j]);
            }
            cout<<endl;
        }*/
        for(int i=1;i<=n;i++){
            for(int j=1;j<=m;j++){
                int now=str[i][j]-'0';
                if(isdigit(str[i][j])){
                    str[i][j]='#';
                    dfs(i,j,now);
                    str[i][j]=now+'0';
                }
            }
        }
        if(flag)cout<<"Possible"<<endl;
        else cout<<"Impossible"<<endl;
    }
    return 0;
}