[递归入门] 走迷宫

题目描述：有一个n*m格的迷宫(表示有n行、m列)，其中有可走的也有不可走的，如果用1表示可以走，0表示不可以走，文件读入这n*m个数据和起始点、结束点(起始点和结束点都是用两个数据来描述的，分别表示这个点的行号和列号)。现在要你编程找出所有可行的道路，要求所走的路中没有重复的点，走时只能是上下左右四个方向。如果一条路都不可行，则输出相应信息(用-l表示无路)。请统一用 左上右下的顺序拓展，也就是 (0,-1),(-1,0),(0,1),(1,0)

输入

第一行是两个数n，m( 1 < n ， m < 15 )，接下来是m行n列由1和0组成的数据，最后两行是起始点和结束点。 

输出

　　所有可行的路径，描述一个点时用(x，y)的形式，除开始点外，其他的都要用“->”表示方向。 
　　如果没有一条可行的路则输出-1。

样例输入

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

样例输出

(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)

 

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <vector>
#include <map>
using namespace std;

int n,m;

const int maxn=20;

int maxtri[maxn][maxn];
int flag[maxn][maxn];

vector<pair<int,int> > ans;

pair<int ,int> begin_point,end_point;

int X[4]={0,-1,0,1};
int Y[4]={-1,0,1,0};

int p_flag=0;

void DFS(int x,int y)
{
    flag[x][y]=1;
    ans.push_back(make_pair(x,y));
    if(x==end_point.first&&y==end_point.second)
    {
        if(p_flag==0)    p_flag=1;
        //ans.push_back(make_pair(x,y));
        for(int i=0;i<ans.size();i++)
        {
            if(i>0)
            {
                printf("->");
            }
            printf("(%d,%d)",ans[i].first,ans[i].second);
        }
        printf("
");
        //ans.pop_back();
        return ;
    }
    for(int i=0;i<4;i++)
    {
        int newx=x+X[i];
        int newy=y+Y[i];
        if(newx>=1&&newx<=n&&newy>=1&&newy<=m&&maxtri[newx][newy]==1&&flag[newx][newy]==0)
        {
            DFS(newx,newy);
            ans.pop_back();
            flag[newx][newy]=0;
        }
    }
}

int main()
{
    memset(flag,0,sizeof(flag));
    scanf("%d %d",&n,&m);
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=m;j++)
        {
            scanf("%d",&maxtri[i][j]);
        }
    }
    //画边界 
    for(int i=0;i<=n+1;i++)
    {
        for(int j=0;j<=m+1;j++)
        {
            if(i==0||j==0||i==n+1||j==m+1)
            {
                maxtri[i][j]=0;
            }
        }
    }
    scanf("%d %d",&begin_point.first,&begin_point.second);
    scanf("%d %d",&end_point.first,&end_point.second);
    DFS(begin_point.first,begin_point.second);
    if(p_flag==0)
    {
        printf("-1
");
    }
    return 0;
}