回溯法也称为试探法,该方法首放弃关于问题规模大小的限制,并将问题的候选解按某一顺序逐一枚举和试验.当发现当前候选解不可能是解时,就选择下一个候选解;倘若当前候选解除了还不满足问题规模要求外,满足所有其他要求时,继续扩大当前候选解的规模,并继续试探.如果当前候选解满足包括问题规模在内的所有要求时,该候选解就是问题的一个解.在回溯法中,放弃当前候选解,寻找下一个候选解的过程称为回溯.扩大当前候选解的规模,并继续试探的过程成为向前试探. 为了确保程序能够终止,调整时,必须保证曾被放弃过的填数序列不被再次试验,即要求按某种有序模型生成填数序列.给解的候选者设定一个被检验的顺序,按这个顺序逐一生成候选者并检验. 对于迷宫问题,我想用回溯法的难点就在如何为解空间排序,以确保曾被放弃过的填数序列不被再次试验.在二维迷宫里面,从出发点开始,每个点按四邻域算,按照右,上,左,下的顺序搜索下一落脚点,有路则进,无路即退,前点再从下一个方向搜索,即可构成一有序模型.下表即迷宫 { 1,1,1,1,1,1,1,1,1,1,
0,0,0,1,0,0,0,1,0,1,
1,1,0,1,0,0,0,1,0,1,
1,0,0,0,0,1,1,0,0,1,
1,0,1,1,1,0,0,0,0,1,
1,0,0,0,1,0,0,0,0,0,
1,0,1,0,0,0,1,0,0,1,
1,0,1,1,1,0,1,1,0,1,
1,1,0,0,0,0,0,0,0,1,
1,1,1,1,1,1,1,1,1,1} 从出发点开始,按序查找下一点所选点列构成有序数列,如果4个方向都搜遍都无路走,就回退,并置前点的方向加1,依此类推....... | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | | x | 1 | 1 | 1 | 2 | 3 | 3 | 3 | 2 | ... | | y | 0 | 1 | 2 | 2 | 2 | 3 | 4 | 4 | ... | | c | 1 | 1 | 3 | 3 | 1 | 1 | 2 | 1 | ... |
#include<stdio.h>
#include<stdlib.h>
#define n1 10
#define n2 10
typedef struct node
{
int x; //存x坐标
int y; //存Y坐标
int c; //存该点可能的下点所在的方向,1表示向右,2向上,3向左,4向右
}linkstack; linkstack top[100];
//迷宫矩阵
int maze[n1][n2]={1,1,1,1,1,1,1,1,1,1,
0,0,0,1,0,0,0,1,0,1,
1,1,0,1,0,0,0,1,0,1,
1,0,0,0,0,1,1,0,0,1,
1,0,1,1,1,0,0,0,0,1,
1,0,0,0,1,0,0,0,0,0,
1,0,1,0,0,0,1,0,0,1,
1,0,1,1,1,0,1,1,0,1,
1,1,0,0,0,0,0,0,0,1,
1,1,1,1,1,1,1,1,1,1,}; int i,j,k,m=0; main()
{
//初始化top[],置所有方向数为左
for(i=0;i<n1*n2;i++)
{
top[i].c=1;
}
printf("the maze is:\n"); //打印原始迷宫矩阵
for(i=0;i<n1;i++)
{
for(j=0;j<n2;j++)
printf(maze[i][j]?"* ":" ");
printf("\n");
}
i=0;top[i].x=1;top[i].y=0;
maze[1][0]=2;
/*回溯算法*/
do
{
if(top[i].c<5) //还可以向前试探
{
if(top[i].x==5 && top[i].y==9) //已找到一个组合
{
//打印路径
printf("The way %d is:\n",m++);
for(j=0;j<=i;j++)
{
printf("(%d,%d)-->",top[j].x,top[j].y);
}
printf("\n"); //打印选出路径的迷宫
for(j=0;j<n1;j++)
{
for(k=0;k<n2;k++)
{
if(maze[j][k]==0) printf(" ");
else if(maze[j][k]==2) printf("O ");
else printf("* ");
}
printf("\n");
} maze[top[i].x][top[i].y]=0;
top[i].c = 1;
i--;
top[i].c += 1;
continue;
}
switch (top[i].c) //向前试探
{
case 1:
{
if(maze[top[i].x][top[i].y+1]==0)
{
i++;
top[i].x=top[i-1].x;
top[i].y=top[i-1].y+1;
maze[top[i].x][top[i].y]=2;
}
else
{
top[i].c += 1;
}
break;
}
case 2:
{
if(maze[top[i].x-1][top[i].y]==0)
{
i++;
top[i].x=top[i-1].x-1;
top[i].y=top[i-1].y;
maze[top[i].x][top[i].y]=2;
}
else
{
top[i].c += 1;
}
break;
}
case 3:
{
if(maze[top[i].x][top[i].y-1]==0)
{
i++;
top[i].x=top[i-1].x;
top[i].y=top[i-1].y-1;
maze[top[i].x][top[i].y]=2;
}
else
{
top[i].c += 1;
}
break;
}
case 4:
{
if(maze[top[i].x+1][top[i].y]==0)
{
i++;
top[i].x=top[i-1].x+1;
top[i].y=top[i-1].y;
maze[top[i].x][top[i].y]=2;
}
else
{
top[i].c += 1;
}
break;
}
}
}
else //回溯
{
if(i==0) return; //已找完所有解
maze[top[i].x][top[i].y]=0;
top[i].c = 1;
i--;
top[i].c += 1;
}
}while(1);
} 
|
本类阅读TOP10