Picking up Jewels

Posted 2020-09-21 yfceshi

Picking up Jewels

There is a maze that has one entrance and one exit. 
Jewels are placed in passages of the maze. 
You want to pick up the jewels after getting into the maze through the entrance and before getting out of it through the exit. 
You want to get as many jewels as possible, but you don’t want to take the same passage you used once.

When locations of a maze and jewels are given, 
find out the greatest number of jewels you can get without taking the same passage twice, and the path taken in this case.

Time limit : 1 sec (Java : 2 sec) (If your program exceeds this time limit, the answers that have been already printed are ignored and the score becomes 0. So, it may be better to print a wrong answer when a specific test case might cause your program to exceed the time limit. One guide for the time limit excess would be the size of the input.)

[Input]
There can be more than one test case in the input file. The first line has T, the number of test cases. 
Then the totally T test cases are provided in the following lines (T ≤ 10 )

In each test case, 
In the first line, the size of the maze N (1 ≤ N ≤ 10) is given. The maze is N×N square-shaped. 
From the second line through N lines, information of the maze is given. 
“0” means a passage, “1” means a wall, and “2” means a location of a jewel. 
The entrance is located on the upper-most left passage and the exit is located on the lower-most right passage. 
There is no case where the path from the entrance to the exit doesn’t exist.

[Output]
For each test case, you should print "Case #T" in the first line where T means the case number.

For each test case, from the first line through N lines, mark the path with 3 and output it.
In N+1 line, output the greatest number of jewels that can be picked up. 

[I/O Example]

Input
2
5
0 0 0 2 0
2 1 0 1 2
0 0 2 2 0
0 1 0 1 2
2 0 0 0 0
6
0 1 2 1 0 0
0 1 0 0 0 1
0 1 2 1 2 1
0 2 0 1 0 2
0 1 0 1 0 1
2 0 2 1 0 0

Output

Case #1

3 0 3 3 3 
3 1 3 1 3 
3 0 3 2 3 
3 1 3 1 3 
3 3 3 0 3 
6

Case #2

3 1 2 1 0 0 
3 1 3 3 3 1 
3 1 3 1 3 1 
3 2 3 1 3 2 
3 1 3 1 3 1 
3 3 3 1 3 3 
4

/*

You should use the statndard input/output

 

in order to receive a score properly.

 

Do not use file input and output

 

Please be very careful.

*/

 

#include <stdio.h>

 

#define MAX_N  11

#define true   1

#define false  0

 

int Array[MAX_N][MAX_N];

int Path[MAX_N][MAX_N];//保存走过的路径

int Visited[MAX_N][MAX_N];//标记走过的路径

int Answer[MAX_N][MAX_N];

int Max_Jewes;

 

 

int N;

//int total ;

//int Answer = 0 ;

 

void SavePath()

{

     int i,j;

     for(i=0;i<N;i++)

     {

         for(j=0;j<N;j++)

         {

              Answer[i][j] = Path[i][j];

         }

     }

}

 

int Visit(int i, int j, int total)

{

     int ret = 0;

     Path[i][j] = 3 ;

     Visited[i][j] = true ;

 

     if((i ==0)&&(j == 0)&&(Array[i][j] == 2))//假设第一个节点为2，total加1

     {

         total++;

     }

    

     if((i == N-1)&&(j == N-1))//假设到达终点，则推断本次遍历的结果和之前全部次数中最大的结果最比較。假设大于之前的结果。则保存这个最大值和本次路径

     {

         if(total > Max_Jewes)

         {

              Max_Jewes = total ;

              SavePath();

         }

        

         Visited[i][j] = false ;//将终点标记成未扫描过，使下一次遍历还能够扫描到该点

         return 1;

     }   

    

     if((j+1) <= (N-1))//down

     {

         if((!Visited[i][j+1])&&(Array[i][j+1] != 1))//假设该点的下方的点没有被扫描过，而且该点下方的点不为墙（1）

         {

              if(Array[i][j+1] == 2)//假设有宝藏，则总数加1。继续遍历

                   Visit(i, j+1, total+1);

              else

                   Visit(i, j+1, total);//假设没有宝藏，则总数不变，继续遍历

 

              Path[i][j+1] = Array[i][j+1] ;//死角，回退之后恢复path的值          

         }

     }

    

     if((j-1) >= 0)//up

     {

         if((!Visited[i][j-1])&&(Array[i][j-1] != 1))

         {

              if(Array[i][j-1] == 2)

                   Visit(i , j-1, total+1);

              else

                   Visit(i , j-1, total);

 

              Path[i][j-1] = Array[i][j-1] ;             

             

         }

     }

 

     if((i+1) <= (N-1))//right

     {

         if((!Visited[i+1][j])&&(Array[i+1][j] != 1))

         {

              if(Array[i+1][j] == 2)

                   Visit(i+1 , j, total+1);

              else

                   Visit(i+1 , j, total);

 

              Path[i+1][j] = Array[i+1][j] ;             

             

         }

     }

    

    

    

     if((i-1) >= 0)//left

     {

         if((!Visited[i-1][j])&&(Array[i-1][j] != 1))

         {

              if(Array[i-1][j] == 2)

                   Visit(i-1 , j, total+1);

              else

                   Visit(i-1 , j, total);

 

             Path[i-1][j] = Array[i-1][j];               

             

         }

     }

     Visited[i][j] = false ;

 

     return -1;

}

 

int main(void)

{

     int T, test_case;

     /*

        The freopen function below opens input.txt file in read only mode, and afterward,

        the program will read from input.txt file instead of standard(keyboard) input.

        To test your program, you may save input data in input.txt file,

        and use freopen function to read from the file when using scanf function.

        You may remove the comment symbols(//) in the below statement and use it.

        But before submission, you must remove the freopen function or rewrite comment symbols(//).

      */

        freopen("sample_input.txt", "r", stdin);

 

     /*

        If you remove the statement below, your program‘s output may not be rocorded

        when your program is terminated after the time limit.

        For safety, please use setbuf(stdout, NULL); statement.

      */

     setbuf(stdout, NULL);

 

     scanf("%d", &T);

     for(test_case = 0; test_case < T; test_case++)

     {

         int i ,j ;

         //total = 0 ;

         Max_Jewes = -1 ;

        

         scanf("%d",&N);

         for(i=0;i<N;i++)

         {

              for(j=0;j<N;j++)

              {

                   scanf("%d",&Array[i][j]);

                   Path[i][j] = Array[i][j];

              }

         }

 

         for(i=0;i<N;i++)

         {

              for(j=0;j<N;j++)

              {

                   Visited[i][j] = false ;////初始化遍历标志

              }

         }

 

        

         Visit(0 , 0, 0);

        

         printf("Case #%d\n", test_case+1);

         //printf("%d\n", Answer);

        

         for(i=0;i<N;i++)

         {

              for(j=0;j<N;j++)

              {

                   printf("%d ",Answer[i][j]);

              }

              printf("\n");

         }       

         printf("%d\n", Max_Jewes);

        

         /////////////////////////////////////////////////////////////////////////////////////////////

         /*

            Implement your algorithm here.

            The answer to the case will be stored in variable Answer.

          */

         /////////////////////////////////////////////////////////////////////////////////////////////

         //Answer = 0;

 

         // Print the answer to standard output(screen).

         //printf("Case #%d\n", test_case+1);

         //printf("%d\n", Answer);

     }

 

     return 0;//Your program should return 0 on normal termination.

}