c ++ Maze

Title Description

A n m lattice maze (expressed n rows, m columns), which can go there not go, if 1 represents go, 0 indicates not go, the file to read in the n m data and from start point, end point (start point and end point are described by two data points represent the row and column number). Now you want to find a way out of all programming viable path, requirements are taking no duplicate points, travel time is only about four directions. If the road are not feasible, then outputs corresponding information (trackless represented by -l).

Please expand unified with the upper left lower right order, that is (0, -1), (- 1,0), (0,1), (1,0)

Entry

The first line is two numbers n, m (1 <n, m <15), followed by n rows and m columns of data 1 and 0 of the last two lines are the start and end points.

Export

All feasible paths, with (x, y) describing a point of the form, in addition to the starting point, the other must use the "->" indicates a direction.

If there is no feasible way to output a -1.

Sample input

5 6
1 0 0 1 0 1
1 1 1 1 1 1
0 0 1 1 1 0
1 1 1 1 1 0
1 1 1 0 1 1
1 1
5 6

Sample Output

(1,1)->(2,1)->(2,2)->(2,3)->(2,4)->(2,5)->(3,5)->(3,4)->(3,3)->(4,3)->(4,4)->(4,5)->(5,5)->(5,6)
(1,1)->(2,1)->(2,2)->(2,3)->(2,4)->(2,5)->(3,5)->(3,4)->(4,4)->(4,5)->(5,5)->(5,6)
(1,1)->(2,1)->(2,2)->(2,3)->(2,4)->(2,5)->(3,5)->(4,5)->(5,5)->(5,6)
(1,1)->(2,1)->(2,2)->(2,3)->(2,4)->(3,4)->(3,3)->(4,3)->(4,4)->(4,5)->(5,5)->(5,6)
(1,1)->(2,1)->(2,2)->(2,3)->(2,4)->(3,4)->(3,5)->(4,5)->(5,5)->(5,6)
(1,1)->(2,1)->(2,2)->(2,3)->(2,4)->(3,4)->(4,4)->(4,5)->(5,5)->(5,6)
(1,1)->(2,1)->(2,2)->(2,3)->(3,3)->(3,4)->(2,4)->(2,5)->(3,5)->(4,5)->(5,5)->(5,6)
(1,1)->(2,1)->(2,2)->(2,3)->(3,3)->(3,4)->(3,5)->(4,5)->(5,5)->(5,6)
(1,1)->(2,1)->(2,2)->(2,3)->(3,3)->(3,4)->(4,4)->(4,5)->(5,5)->(5,6)
(1,1)->(2,1)->(2,2)->(2,3)->(3,3)->(4,3)->(4,4)->(3,4)->(2,4)->(2,5)->(3,5)->(4,5)->(5,5)->(5,6)
(1,1)->(2,1)->(2,2)->(2,3)->(3,3)->(4,3)->(4,4)->(3,4)->(3,5)->(4,5)->(5,5)->(5,6)
(1,1)->(2,1)->(2,2)->(2,3)->(3,3)->(4,3)->(4,4)->(4,5)->(5,5)->(5,6)

Source Code

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

int n,m,a[20][20],sx,sy,tx,ty,ans[300][2],used[20][20],opt[20][20],flag;
int d[4][2]={{0,-1},
             {-1,0},
             {0,1},
             {1,0}};
void dfs(int x,int y,int step)//step是步数
{
    if(x == tx && y == ty)//如果起点等于终点
    {
        memset(opt,0,sizeof(opt));//memset把opt[]清零
        for(int i = 1;i <= step;i ++)//循环输出步数
            if(opt[ans[i][0]][ans[i][1]] == 1) return ;//这是一条可行路线
            else opt[ans[i][0]][ans[i][1]] = 1;
        printf("(%d,%d)",ans[1][0],ans[1][1]);
        for(int i = 2;i <= step;i ++)//循环输出除了起点的所有路线
            printf("->(%d,%d)",ans[i][0],ans[i][1]);//ans[2][0]输出第2步的x坐标ans[2][1]输出第2步的y坐标
        printf("\n");
        flag = 1;//flag = 1表示已经输出过(有可行的路线)
        return ;
    }
    for(int i=0;i<4;i++)
    {
        int vx = x + d[i][0],vy = y + d[i][1];//上下左右扩展
        if(vx < 1 || vx > n || vy < 1 || vy > m) continue;//如果在迷宫外面
        if(a[vx][vy] == 0) continue;//如果不能走
        if(used[vx][vy] == 1) continue;//如果走过
        used[vx][vy] = 1;//标位已经走过 作用:不能重复走
//        printf("%d   %d   %d\n\n",vx,vy,step + 1);
        ans[step + 1][0] = vx;ans[step + 1][1] = vy;//ans[2][0]输出第2步的x坐标ans[2][1]输出第2步的y坐标
        dfs(vx,vy,step + 1);//递归寻找下一个可以走的节点(不断递归下一个节点是否可行,如果下个节点B的其中两个方向被堵死,第三个方向C虽然能走,但是C被堵死 也删除下个节点B)
        
        used[vx][vy] = 0;
    }
}
int main()
{
    std::cin >> n >> m;
    for(int i = 1;i <= n;i ++)
        for(int j = 1;j <= m;j ++)
            std::cin >> a[i][j];//矩阵转换成数组
    std::cin >> sx >> sy;//起点
    std::cin >> tx >> ty;//终点
    ans[1][0] = sx;//ans[1][0]:1表示步数  0表示存的是x坐标
    ans[1][1] = sy;//ans[1][0]:1表示步数  1表示存的是y坐标
    dfs(sx,sy,1);//从起点开始递归查找
    if(flag == 0)
        printf("-1");
    return 0;
}
/*
 input:
 
 
5 6
1 0 0 1 0 1
1 1 1 1 1 1
0 0 1 1 1 0
1 1 1 1 1 0
1 1 1 0 1 1
1 1
5 6
 
*/