HDU 5612 Baby Ming and Matrix games（dfs暴力）

Baby Ming and Matrix games

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1403    Accepted Submission(s): 382

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

大体题意：
给你个n*n的字符串。
n一定是奇数，其中包括数字（在奇数行或者奇数列）
包括运算符号（偶数行或者偶数列）
还有包括没用的字符#  不应该计算！

思路：

dfs 暴力搜索：
遍历字符串，发现一个数字直接dfs。
在dfs中，判断和sum相等  直接return true；（要加精度控制，因为涉及除法，用double ，就得加eps）
然后4个方向走两个单位，因为符号是走一个单位，当走两个单位合法时，那么一个单位走到符号位也是合法的！
然后在dfs传参数即可！
在一个就是除法时，要先判断分母是否为0，是0 则不计算！


教训：
数组大小一定要合理，这个题目数组开小，竟然是WA~~T_T!!
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cstdlib>
#include<cmath>
#include<cctype>
using namespace std;
int n,m;
double sum;
const int maxn = 105;
const double eps = 1e-8;
bool vis[maxn][maxn];
char s[maxn][maxn];
const int dx[] = {2,0,-2,0};
const int dy[] = {0,-2,0,2};
void init(){
    memset(vis,0,sizeof vis);
}
bool dfs(int x,int y,double ans){
    if (fabs(ans-sum) < eps)return true;
    for (int i = 0; i < 4; ++i){
        int xx = x + dx[i];
        int yy = y + dy[i];
        int x2 = x + dx[i]/2;
        int y2 = y + dy[i]/2;
        if (xx >= 1 && xx <= n && yy >= 1 && yy <= m && !vis[xx][yy] && s[xx][yy] != '#'){
            if (s[x2][y2] == '+'){
                vis[xx][yy] = 1;
                if (dfs(xx,yy,ans + s[xx][yy]-48))return true;
                vis[xx][yy] = 0;
            }
            if (s[x2][y2] == '-'){
                vis[xx][yy] = 1;
                if (dfs(xx,yy,ans - (s[xx][yy]-48)))return true;
                vis[xx][yy] = 0;
            }
            if (s[x2][y2] == '*'){
                vis[xx][yy] = 1;
                if (dfs(xx,yy,ans * (s[xx][yy]-48)))return true;
                vis[xx][yy] = 0;
            }
            if (s[x2][y2] == '/' && s[xx][yy]!=48){
                vis[xx][yy] = 1;
                if (dfs(xx,yy,ans / (s[xx][yy]-48)))return true;
                vis[xx][yy] = 0;
            }
        }
    }
    return false;

}
int main(){
    int T;
    scanf("%d",&T);
    while(T--){
        init();
        scanf("%d%d%lf",&n,&m,&sum);
        for (int i = 1; i <= n; ++i)
            scanf("%s",s[i] + 1);
        bool ok = false;
        for (int i = 1; i <= n; ++i){
            for (int j = 1; j <= m; ++j){
                if (isdigit(s[i][j])){
                    memset(vis,0,sizeof vis);
                    vis[i][j] = 1;
                    if (dfs(i,j,s[i][j]-48)){
                        ok=true;
                        goto TT;
                    }
                }
            }
        }
        TT:
            printf("%s",ok ? "Possible" : "Impossible");
            printf("\n");
    }
    return 0;
}