[英]Moving a sprite to the left or down is faster than moving it to the right or up
[英]Moving down, right, left and up in matrix recursively causes stack overflow exception
該程序用於找出矩陣中兩點之間的最短路徑,其中我向下,向右,向左和向上遍歷,但由於遞歸,它進入一個來回的無限循環。
這個程序基本上遍歷矩陣所在
問題是找到B和C之間最短的浴槽。
我怎么能讓這段代碼工作? 如同在一次之后停止控制向下。
import java.util.*;
import java.lang.*;
import java.io.*;
/* Name of the class has to be "Main" only if the class is public. */
class Stockroom
{
//static int m = 0;
//static int n = 0;
//static char a[][] = new char [m][n];
public static boolean checkFeasibility(int x, int y, int row, int col, char a[][])
{
if(x>=0 && x<row && y>=0 && y<col && a[x][y] != 'D')
return true;
else
return false;
}
public static boolean shortestPath(char a[][], int bx, int by, int x, int y, int len, int minLen)
{
if( checkFeasibility(bx,by,x,y,a)==false )
return false;
if(a[bx][by]=='C')
{
minLen = Math.min(len,minLen);
System.out.println(minLen-1);
return true;
}
len++;
if(shortestPath(a,bx+1,by,x,y,len++,minLen)== true)
return true;
if(shortestPath(a,bx,by+1,x,y,len++,minLen)==true)
return true;
if(shortestPath(a,bx,by-1,x,y,len++,minLen)== true)
return true;
if(shortestPath(a,bx-1,by,x,y,len++,minLen)== true)
return true;
else {
len--;
return false;
}
}
public static void main (String[] args) throws java.lang.Exception
{
char arr[][] = {
{'_','B','_','_'},
{'D','_','_','D'},
{'_','D','_','_'},
{'_','_','C','_'},
};
int bx =0,by=1,px=3,py=2;
int n =4,m=4;
shortestPath(arr, bx, by, m, n, 0, 100);
}
}
只需使用“D”標記您訪問過的每個字段,以避免返回。 因此,在shortestPath
,打完電話后checkFeasibility
和檢查后,如果該值是“C”,這樣做:
a[bx][by] = 'D';
詳細闡述了Frank Puffer的想法:
class Stockroom {
public static boolean checkFeasibility(int x, int y, int row, int col,
char a[][]) {
if (x >= 0 && x < row && y >= 0 && y < col && a[x][y] != 'D')
return true;
else
return false;
}
public static boolean shortestPath(char a[][], int bx, int by, int x,
int y, int len, int minLen) {
if (checkFeasibility(bx, by, x, y, a) == false)
return false;
if (a[bx][by] == 'C') {
minLen = Math.min(len, minLen);
System.out.println(minLen - 1);
return true;
}
len++;
if (len >= minLen) { // this was not shortest
return false;
}
// hack to make sure we don’t go through the same spot again
a[bx][by] = 'D';
if (shortestPath(a, bx + 1, by, x, y, len, minLen) == true) {
// remove temporary block so this space can be used in other paths
a[bx][by] = '_';
return true;
}
if (shortestPath(a, bx, by + 1, x, y, len, minLen) == true) {
a[bx][by] = '_';
return true;
}
if (shortestPath(a, bx, by - 1, x, y, len, minLen) == true) {
a[bx][by] = '_';
return true;
}
if (shortestPath(a, bx - 1, by, x, y, len, minLen) == true) {
a[bx][by] = '_';
return true;
}
len--;
return false;
}
public static void main(String[] args) {
// find path from B to C; don’t go through D
char arr[][] = { { '_', 'B', '_', '_' },
{ 'D', '_', '_', 'D' },
{ '_', 'D', '_', '_' },
{ '_', '_', 'C', '_' },
};
int bx = 0, by = 1, px = 3, py = 2;
int n = 4, m = 4;
shortestPath(arr, bx, by, m, n, 0, 100);
System.out.println(Arrays.deepToString(arr));
}
}
這會修復'_'字段的覆蓋,但仍會覆蓋'B'。 程序打印3,因為最短路徑長度為4,您減去1。
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