c - Futoshiki C递归求解器

Question

所以我有这个程序可以解决 C 中的 futoshiki 难题这是从具有以下格式的文本文件加载的:

50 | 0 | 0 | 0 | 0- - - - v - - - - 0 > 0 | 0 | 0 | 3- - - - - - - - - 0 | 0 < 2 | 0 | 0- - - - v - - - - 0 | 0 | 0 | 0 | 4^ - v - - - - - - 0 | 0 | 0 | 0 | 0

where 5 is the size of the matrix, and the numbers adjcent to the operators <, >, ^, v must satisfy the condition imposed by them, from the file all the characters on rows are divided by spaces eg 0 |...So I've managed to load the file, to check if it satisfies the math operators conditions, but I'm stuck on the recursive function

What I'd like to know:

Did i choose the right way to store the matrix or I've should have divided the numbers from the logical operators ?

How could I perform an recursive expansion on the matrix and how could I track the used number in a certain step(in case I would have to backtrack)?

eg. let's say I arrive at index[j][j] where j<n (size of matrix) , starting from there I would have to decrement j ("touching") only numbers and check if the sub-matrix satisfies the conditions

Here's what I've managed to code so far.

where :

char **readmat(int *n); //reads the matrix from the file eliminating the spaces between chars

void print(char **mat,int n); //prints the stored matrix

int check(char **mat,int n); //checks if items of a matrix of size n satisfies the math operators

int expand (char **mat,int n,int i); //this should be the recursive functions that gets an element at a time and checks if there's any condition to be satisfied, if so, increments it

#include <stdio.h>
#include <stdlib.h>
#include <string.h>


char **readmat(int *n);
void print(char **mat,int n);
int check(char **mat,int n); 
int expand (char **mat,int n,int i);

int main(int argc, char *argv[])
  {
  char **mat;
  int n, j;

  mat=readmat(&n);

  if(mat == NULL)
     return 1;

  if(check(mat,n)){
     print(mat,n);
  }
  else if(expand(mat,n,0)==1){
      print(mat,n);
  }
  else {
      printf("Nessuna soluzione trovata.\n");
  }

  for(j=0; j<=n;j++)
       free(mat[j]);
  free(mat);

  system("PAUSE");  
  return 0;
}

char **readmat(int *n){
     FILE *fp;
     char *line,nome[100];
     int i,j,k;
     char **mat;

     printf("Inserire il nome del file: ");
     scanf("%s",nome);
     fp=fopen(nome,"r");
     if(fp==NULL){
     printf("Errore apertura file");
     return NULL;
 }

 if(fgets(nome,100,fp)==NULL){
     printf("Formato file non valido\n");
     fclose(fp);
     return NULL;
 }
 if(sscanf(nome,"%d",n)!=1){
     printf("Errore nei parametri del file\n");
     fclose(fp);
     return NULL;    
 }

 (*n)=(((*n)*2)-1);


 mat=(char**)malloc((*n)*sizeof(char*));
 for(i=0;i<=(*n);i++)
    mat[i]=(char*)malloc((*n)*sizeof(char));

 line=(char*)malloc(2*(*n)*sizeof(char));

 i=0;

 while(i<=2*(*n) && fgets(line,2*(*n)+2,fp)!=NULL){
    j=0;
    k=0;
    while(j<=2*(*n)){
        if(line[j]!=' '){
           mat[i][k]=line[j];
           k++;
        }  
        j++;
    }
    i++;
 }   
 return mat;
 //print(mat, (*n));  
}

void print(char **mat,int n){
    int i=0,j=0;
    for (i=0; i<n; i++) {
     for (j=0; j<n; j++) {
        printf("%c", mat[i][j]);
     }
     printf("\n");
    }
}

int check(char **mat,int n) {

    int i,j;
    int k=1;

    for(i=0;i<n;i++){
        for(j=0;j<n;j++){
            if(mat[i][j]=='<'){
                if(mat[i][j-1] >= mat[i][j+1])
                    k=0;                               
            }
            else if(mat[i][j]=='>'){
                if(mat[i][j-1] <= mat[i][j+1])
                    k=0;
            }   
            else if(mat[i][j]=='^'){
                if(mat[i-1][j] >= mat[i+1][j])
                    k=0;    
            }
            else if(mat[i][j]=='v'){
                if(mat[i-1][j] <= mat[i+1][j])
                    k=0;                      
            }                            
        }
    }
    return k;                                
}
int expand (char **mat,int n,int i){

    int j=i/n;
    int k=i%n;
    int p;

    if(i>=n*n){

        return 1;
    }       
    else{
        if((mat[j][k]>47)&&(mat[j][k]<58)){
            if(mat[j][k]=='0'){     
                expand(mat,n,i+2);
            }   
            for (p=(mat[j][k]-48); p<(10-(mat[j][k]-48)); p++) {              
                mat[j][k]=48+p;                        
                if (check(mat,i)) {
              if (expand(mat, n, i+2)) {
                   return 1;
                    }
                }
            }
            i-=2;
            mat[j][k]='0';
        }
    }           
    return 0;
}

示例的解决方案:如您所见，逻辑条件区域显然得到满足

0 | 0 | 1 | 0 | 0
- - - - v - - - - 
1 > 0 | 0 | 0 | 3
- - - - - - - - - 
0 | 0 < 2 | 0 | 0
- - - - v - - - - 
0 | 1 | 0 | 0 | 4
^ - v - - - - - - 
1 | 0 | 0 | 0 | 0

Answer 1

存储矩阵的方式无关紧要。可以任意存储，只要能方便地获取/设置每个点的数值，并评估操作者是否满意即可。

从广义上讲，您可以使用如下算法解决此类问题:

//returns true if this function solved the puzzle, false otherwise.
//gameData will be changed to the solved form of the puzzle if a solution exists, or remain unchanged if no solution exists.
//(so, whatever language you're using, ensure gameData is passed by reference here)
bool solve(gameData){
    if (!isValid(gameData)){return false;}  //oops, puzzle is unsolvable!
    if (isComplete(gameData)){return true;} //puzzle is already solved; no further modification needed.

    //choose a spot on the game board that hasn't been filled in yet.
    int x;
    int y;
    getEmptySpot(gameData, &x, &y);

    //iterate through all the possible values that could go into the empty spot.
    //you don't need anything fancy here to generate legal values for i;
    //if you accidentally supply an invalid value, then isValid()
    //will notice in the next solve() call.
    for (int i = 1; i <= 5; i++){
        //try putting i in the empty spot.
        setValue(gameData, x, y, i);
        if (solve(gameData)){ //putting i in the spot led to a solution!
            return true;
        }
    }
    //didn't find a solution :(
    //return gameData to its original state.
    setValue(gameData, x, y, 0);
    return false;
}

该算法进行强力递归搜索，为每个点尝试每个可能的值，如果进入非法状态则回溯。在 super 最坏的情况下，它以指数时间运行，但实际上，开始时的 isValid() 调用将短路任何明显不可行的分支，因此它应该以 5x5 的速度相当快地完成输入。

isValid、isComplete、getEmptySpot 和 setValue 的实现将取决于您如何定义 gameData。

isValid 应该检查游戏数据是否处于非法状态 - 在您的情况下，它应该检查所有大于比较是否正确，并检查每个数字是否出现每行和每列仅一次。这些检查应忽略值为 0 的点，因为它们只是一个占位符，表示“尚未填充”。

isComplete 应检查以查看没有任何点具有“尚未填充”占位符。 (isValid(gameData) && isComplete(gameData)) 表示 gameData 已解决。

getEmptySpot 应该找到尚未填充的位置。如果您担心速度，它应该找到可以合法输入的最小数值的位置。这将大大减少搜索树的宽度。

最后，setValue 应该将给定点设置为给定值。