Hdu 5016 Baby Ming and Matrix games【dfs】

Baby Ming and Matrix games

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

Total Submission(s): 1207 Accepted Submission(s): 314

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($-10^{18} < sum < 10^{18}$, 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 $15$) 

$1 \leq T \leq 1000$
 

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：

一幅图，给出规模和目标值，问能否通过某些数字和符号的组合得到目标的值

.

题解：

直接暴力搜索+回溯

也许今生都忘不了这道题！！！

错了不知道几十遍，从三点多调到现在，终于敢写博客了.....

本来是一个简单的dfs加上回溯，结果由于自己的一些小疏忽，一直在细节上wa1...最终还是比较完善的解决掉了....

另外精度的控制，真的不知道要怎么办了，到底多少才是适合的？？？

#include<stdio.h>
#include<string.h>
#include<math.h>
int n,m,kase,vis[55][55],dx[4]={0,0,-2,2},dy[4]={-2,2,0,0};
char map[55][55];
double s;
double num(double x,double y,char ch)
{
switch (ch)
{
case '+':return x+y;
case '-':return x-y;
case '*':return x*y;
case '/':return x/y;
}
}
void dfs(int x,int y,double sum)
{
if(kase)//找到结果了
{
return;
}
if(fabs(sum-s)<1e-6)//相等了
{
kase=1;
return;
}
for(int i=0;i<4;++i)
{
int tx=x+dx[i],ty=y+dy[i];
if(tx<0||tx>=n||ty<0||ty>=m)//越界
{
continue;
}
int tp=map[tx][ty]-'0';
char op=map[x+dx[i]/2][y+dy[i]/2];
if(tp==0&&op=='/')//除零错误
{
continue;
}
if(!vis[tx][ty])
{
vis[tx][ty]=1;
dfs(tx,ty,num(sum,tp,op));
vis[tx][ty]=0;//回溯
}
}
}
void slove()
{
kase=0;
memset(vis,0,sizeof(vis));
for(int i=0;i<n;i+=2)
{
for(int j=0;j<m;j+=2)
{
<span style="white-space:pre"> </span>vis[i][j]=1;
dfs(i,j,map[i][j]-'0');
if(kase)
{
printf("Possible\n");//符合要求了
return;
}
vis[i][j]=0;//取消标记
}
}
printf("Impossible\n");
}
int main()
{
int t;
//freopen("shuju.txt","r",stdin);
scanf("%d",&t);
while(t--)
{
scanf("%d%d%lf",&n,&m,&s);
for(int i=0;i<n;++i)
{
scanf("%s",map[i]);
}
slove();
}
return 0;
}