HDU 5612 Baby Ming and Matrix games (搜索)-优快云博客

在给定的矩阵中，从任意数字开始，通过相邻数字的运算，尝试找到一种方式，使得最终结果等于指定的整数。通过遍历和递归搜索矩阵中的数字和运算符，实现对矩阵游戏的挑战。

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Baby Ming and Matrix games

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 392 Accepted Submission(s): 82

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：输入n,m,sum分别代表着矩阵的长，宽，要求的值，从矩阵的任何一个位置是0~9的数开始搜索，能否找到一个连续的计算式，使这个计算式的值为sum

方法：暴搜！！！！

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cmath>

#define LL long long
#define eps 1e-10

using namespace std;
double sum;
int n,m;
char str[50][50];
bool vis[55][55];
int Dir[][2] = {{0,1},{0,-1},{1,0},{-1,0}};

int DFS(int x, int y, double ant)
{

    if(fabs(ant - sum) <= eps)
        return 1;
    vis[x][y] = true;
    for(int i=0; i<4; i++)
    {
        int xx = x + Dir[i][0]*2;
        int yy = y + Dir[i][1]*2;
        int Deno = str[xx][yy] - '0';
        if(vis[xx][yy]|| xx < 0 || xx >= n || yy < 0 || yy >= m)
            continue;
        else if(str[x + Dir[i][0]][y + Dir[i][1]] == '+')
        {
            if(DFS(xx,yy,ant+Deno))
                return 1;
        }
        else if(str[x + Dir[i][0]][y + Dir[i][1]] == '-')
        {
            if(DFS(xx,yy,ant-Deno))
                return 1;
        }
        else if(str[x + Dir[i][0]][y + Dir[i][1]] == '*')
        {
            if(DFS(xx,yy,ant*Deno))
                return 1;
        }
        else if(str[x + Dir[i][0]][y + Dir[i][1]] == '/')
        {
            if(Deno)
            {
                if(DFS(xx,yy,ant/Deno))
                    return 1;
            }
        }
    }
    vis[x][y] = false;
    return 0;
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T-- && scanf("%d%d%lf",&n,&m, &sum))
    {
        memset(vis, false, sizeof(vis));
        for(int i=0; i<n; i++)
            scanf("%s",str[i]);
        int flag = 0;
        for(int i=0; i<n&&!flag; i+=2)
            for(int j=0; j<m&&!flag; j+=2)
                flag |= DFS(i,j,str[i][j]-'0');

        if(flag)
            printf("Possible\n");
        else
            printf("Impossible\n");
    }


    return 0;
}