hdoj 1010 <dfs剪枝>

Tempter of the Bone

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 106662    Accepted Submission(s): 29000

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

 

Author

ZHANG, Zheng

 

Source

ZJCPC2004

做的要疯了，最后看了别人的博客，才发现不是剪枝没剪好---而是自己以前写dfs 从来没有注意的一个坏习惯-： 找到满足条件后--其他dfs应该一律跳出------

代码：

#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>using namespace std;int n,m,k,kx,ky,jx,jy;bool fa,fafe[8][8];char ma[8][8];int zx[4]={0,0,1,-1};int zy[4]={1,-1,0,0};void dfs(int xx,int yy,int s){    if (fa)        return ;    if (xx==jx&&yy==jy)    {        if (s==k)        {            fa=true;            return ;        }        else        return ;    }    if (abs(jx-xx)+abs(jy-yy)+s>k) return ;    for (int i=0;i<4;i++)    {       int  px=xx+zx[i];int py=yy+zy[i];       if (px<0||py<0||px==n||py==m||ma[px][py]=='X')       continue ;       ma[px][py]='X';       dfs(px,py,s+1);       ma[px][py]='.';    }}int main(){    while (scanf("%d%d%d",&n,&m,&k),n+m+k)    {        for (int i=0;i<n;i++)            scanf("%s",ma[i]);            int qiang=0,op=0;        for (int i=0;i<n;i++)        {            for (int j=0;j<m;j++)            {                if (ma[i][j]=='S')                {                    kx=i,ky=j;                }                else if (ma[i][j]=='D')                {                    jx=i;jy=j;                    op=1;                }                else if (ma[i][j]=='X')                    qiang++;            }        }        if (op==0)            printf("NO\n");        else if (n*m-qiang<k||(abs(jx-kx)+abs(jy-ky))%2!=k%2||abs(jx-kx)+abs(jy-ky)>k||k>=n*m)        {            printf("NO\n");        }        else        {            fa=false;            memset(fafe,true,sizeof(fafe));            ma[kx][ky]='X';            dfs(kx,ky,0);            if (fa)                printf("YES\n");            else                printf("NO\n");        }    }    return 0;}