hdu 5612 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): 1257    Accepted Submission(s): 330

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 
    
    
    
    
     
   
   
   
   

 

Source

BestCoder Round #69 (div.2)

 

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深搜找出所有答案，只是注意浮点数的比较问题；

#include<iostream>
#include<cstdlib>
#include<string>
#include<algorithm>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<stack>
#include<queue>
#include<iomanip>
#include<map>
#include<set>
#include<functional>
#define pi 3.14159265358979323846
using namespace std;
int dx[]={1,-1,0,0};
int dy[]={0,0,1,-1};
int n,m;
double sum;
bool vis[30][30];
char maze[30][30];
bool flag;
void dfs(int x,int y,double cursum)//cursum为当前的结果
{
    if(fabs(cursum-sum)<1e-7)
    {
        flag=1;
        return ;
    }
    if(flag==1)
    {
        return ;
    }
    for(int k=0;k<4;++k)
    {
        int tx=x+dx[k]*2,ty=y+dy[k]*2;
        if(tx<1||tx>n||ty<1||ty>m||vis[tx][ty])
            continue;
        char cal=maze[x+dx[k]][y+dy[k]];
        int tnum=maze[tx][ty]-'0';
        vis[tx][ty]=1;
        if(cal=='+') dfs(tx,ty,cursum+tnum);
        if(cal=='-') dfs(tx,ty,cursum-tnum);
        if(cal=='*') dfs(tx,ty,cursum*tnum);
        if(cal=='/') dfs(tx,ty,cursum/tnum);
        vis[tx][ty]=0;
    }
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        flag=0;
        memset(vis,0,sizeof(vis));
        scanf("%d %d %lf",&n,&m,&sum);
        for(int i=1;i<=n;++i)
        {
            scanf("%s",maze[i]+1);
        }
        for(int i=1;i<=n;i+=2)
            for(int j=1;j<=m;j+=2)
        {
            vis[i][j]=1;
            dfs(i,j,maze[i][j]-'0');
            vis[i][j]=0;
            if(flag) break;
        }
        if(flag)
            puts("Possible");
        else
            puts("Impossible");
    }
    return 0;
}