利用dfs解决规定路程问题

今天继续dfs的训练，遇到了一道神题，不停地TLE，我DD都快碎了。。。。。好在经过本渣不懈努力，还是弄了出来，不容易啊，发上来纪念一下，顺便总结一下关于用dfs解决规定路程的问题。

先上题目：

Description

The doggie found a bone in an ancient maze, which fascinated him a lot. However, when he picked it up, the maze began to shake,

 and the doggie could feel the ground sinking. He realized that the bone was a trap, and he tried desperately to get out of this maze. 



The maze was a rectangle with sizes N by M. There was a door in the maze. At the beginning, the door was closed and it would 

open at the T-th second for a short period of time (less than 1 second). Therefore the doggie had to arrive at the door on exactly

 the T-th second. In every second, he could move one block to one of the upper, lower, left and right neighboring blocks. Once 

he entered a block, the ground of this block would start to sink and disappear in the next second. He could not stay at one block 

for more than one second, nor could he move into a visited block. Can the poor doggie survive? Please help him. 

 

Input

The input consists of multiple test cases. The first line of each test case contains three integers N, M, and T (1 < N, M < 7; 

0 < T < 50), which denote the sizes of the maze and the time at which the door will open, respectively. The next N lines give

 the maze layout, with each line containing M characters. A character is one of the following: 



'X': a block of wall, which the doggie cannot enter; 

'S': the start point of the doggie; 

'D': the Door; or 

'.': an empty block. 



The input is terminated with three 0's. This test case is not to be processed. 

 

Output

For each test case, print in one line "YES" if the doggie can survive, or "NO" otherwise. 

 

Sample Input

 4 4 5
S.X.
..X.
..XD
....
3 4 5
S.X.
..X.
...D
0 0 0 

 

Sample Output

 NO
YES 

 

附上本渣题解：

//基本思路：通过回溯（即return回来后）把标记的改为未标记的从而得到所有路径
#include<stdio.h>
#include<cstdlib>
int r,c,t,x,y,x1,y1;
int dx[4]={0,0,1,-1};
int dy[4]={1,-1,0,0};
char point[7][7];
bool flag;
int count;
int map[7][7];//能够帮助去掉不可能的情况，不用就TLE
void dfs(int x0,int y0,int r,int c,int step)//step记录当前点在当前路径的步数
{
    int temp;
    temp=count-step-abs(x0-x1)-abs(y0-y1)+1;//通过对当前点和最终点的位置判断当前点有无可能满足条件，不用就TLE
    if (temp<0||temp%2==1)//这是通过奇偶判断，数学知识，易忽视。。
        return;
    for(int i=0;i<4;i++)
    {
        int tempx=x0+dx[i],tempy=y0+dy[i];
        if(point[tempy][tempx]=='D'&&step==count)//判断步数与路径长度是否相同
            flag=true;
        if(tempx>=0&&tempx<c&&tempy>=0&&tempy<r&&point[tempy][tempx]=='.')
        {
            point[tempy][tempx]='X';//通过使当前点变为X来标记当前路径是否经过
            dfs(tempx,tempy,r,c,step+1);//探索邻接点
            point[tempy][tempx]='.';//回溯还原，使其得能够探索全部路径
            if(flag==true)//如果存在，就直接返回，不用此判断就TLE。。。
                return ;
        }
    }
    return ;
}
int main()
{
    for(int i=0;i<7;i++)
        for(int j=0;j<7;j++)
        {
            if((i+j)%2)
                map[i][j]=0;
            else
                map[i][j]=1;
        }
    while(1)
    {
        scanf("%d%d%d",&r,&c,&t);
        getchar();
        if(r==0&&c==0&&t==0)
            break;
        count=t;
        for(int i=0;i<r;i++)
        {
            scanf("%s",point[i]);
        }
        for(int i=0;i<r;i++)
            for(int j=0;j<c;j++)
            {
                if(point[i][j]=='S')
                {
                    y=i;
                    x=j;
                }
                else if(point[i][j]=='D')
                {
                    y1=i;
                    x1=j;
                }
            }
        if(abs(map[y1][x1]-map[y][x])%2!=t%2)//map判断起点终点的位置是否有可能，不用就TLE。。
        {
            printf("NO\n");
            continue;
        }
        flag=false;
        dfs(x,y,r,c,1);
        if(flag)
            printf("YES\n");
        else
            printf("NO\n");
    }
    return 0;
}

map的用处：

v可以把map看成这样：

v0 1 0 1 0 1

v1 0 1 0 1 0

v0 1 0 1 0 1

v1 0 1 0 1 0

v0 1 0 1 0 1

v从为 0的格子走一步，必然走向为
1的格子

v从为 1的格子走一步，必然走向为
0的格子

v即：

v0->1或1->0必然是奇数步

v0->0走1->1必然是偶数步 

所以当遇到从 0走向0或从1走向1但是要求时间是奇数的，或者，从1
走向 0
但是要求时间是偶数的都可以直接判断不可达！

所以map表能从一开始就减少可能的情况。