hdoj 1010 Tempter of the Bone (dfs 奇偶剪枝)

Tempter of the Bone

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 107079    Accepted Submission(s): 29114

Problem 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 5S.X...X...XD....3 4 5S.X...X....D0 0 0

 

Sample Output

NOYES

 

代码：

#include <iostream>#include <cstdio>#include <cstring>#include<cmath>using namespace std;int vis[11][11];char map[11][11];int dir[4][2]={0,1,0,-1,1,0,-1,0};int n,m,t;int cnt;int sx,sy,dx,dy;int flag;int judge(int x,int y){if(map[x][y]!='X'&&vis[x][y]==0&&x>=0&&x<n&&y>=0&&y<m)return 1;return 0;}void dfs(int x,int y,int cnt){if(flag)//  (这里不加，就会超时)//这里是告诉已经找到，结束其他上层的dfs return;if(x==dx&&y==dy&&cnt==t)//这里只是结束当前的dfs {flag=1;return;}int temp=t-cnt-(abs(dx-x)+abs(dy-y));//剪枝 if(temp<0||temp&1)return;for(int i=0;i<4;i++){int nx=x+dir[i][0];int ny=y+dir[i][1];if(judge(nx,ny)){vis[nx][ny]=1;dfs(nx,ny,cnt+1);vis[nx][ny]=0;}}}int main(){    while(~scanf("%d%d%d",&n,&m,&t),n+m+t)    {        int num=0;//        getchar();        for(int i=0;i<n;i++)        {            scanf("%s",map[i]);            for(int j=0;j<m;j++)            {                if(map[i][j]=='S')                {                    sx=i;                    sy=j;                }                if(map[i][j]=='D')                {                    dx=i;                    dy=j;                }                if(map[i][j]=='.')                {                    num++;                }            }//            getchar();        }        if(num+1<t)//可以走的路小于时间         {            printf("NO\n");            continue;        }        flag=0;        memset(vis,0,sizeof(vis));        vis[sx][sy]=1;               dfs(sx,sy,0);        if(flag)            printf("YES\n");        else            printf("NO\n");    }    return 0;}

• 最重要的根据曼哈顿距离（从一个点到达另外一个点的最短路径长度|x1-x2|+|y1-y2|）进行剪枝，路径长度(非最短)与最短路径的长度同奇偶,它们的差一定是偶数。所以判断当前位置到终点的曼哈顿距离与它到终点如果成立所需要的路径长度的奇偶性，如果它们的差是奇数则可以停止递归。

奇偶剪枝：

是数据结构的搜索中，剪枝的一种特殊小技巧。

现假设起点为(sx,sy)，终点为（ex,ey），给定t步恰好走到终点，

　

s    |    |    |    +———e

 

如图所示（“|”竖走，“—”横走，“+”转弯），易证abs(ex-sx)+abs(ey-sy)为此问题类中任意情况下，起点到终点的最短步数，记做step，此处step1=8；

　　

s———  ——+ |+   |    +———e

 

如图，为一般情况下非最短路径的任意走法举例，step2=14；

step2-step1=6，偏移路径为6，偶数（易证）；

故，若t-[abs(ex-sx)+abs(ey-sy)]结果为非偶数（奇数），则无法在t步恰好到达；

返回，false；

反之亦反。