HDU 1010Tempter of the Bone(奇偶剪枝回溯dfs)

Tempter of the BoneTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 55541    Accepted Submission(s): 14983


Problem DescriptionThe 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.
 
InputThe 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.
 
OutputFor 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

 
AuthorZHANG, Zheng 
SourceZJCPC2004 

      题目大意：要求每个点只能走一次，从起点走到终点刚好sum步，点表示可以走得路经。S，D分别代表起点终点。
解题思路：每个点只能走一次，所以必须回溯。开始没想到剪枝，果断TLE了，而且因为yes,no的大小写还WA了一发脑残了。
        题目地址：Tempter of the Bone
AC代码：

#include<iostream>#include<cstring>#include<string>#include<cmath>#include<cstdio>using namespace std;char a[10][10];int visi[10][10];int n,m,sum,flag;struct node{    int x;    int y;};node sta,en;  //起点与终点int dir[4][2]=  //四个方向{    {-1,0},{1,0},{0,-1},{0,1}};void dfs(int cx,int cy,int step){    int i;    if(flag==1) return;   //找到解    if(step==sum)    {        if(cx==en.x&&cy==en.y)           flag=1;        return;    }    int need=abs(cx-en.x)+abs(cy-en.y);    int tmp=sum-step;    if((tmp%2==0&&need%2==1)||(need%2==0&&tmp%2==1)||tmp<need)  //奇偶剪枝       return;    for(i=0;i<4;i++)    {        int px,py;        px=cx+dir[i][0],py=cy+dir[i][1];        if(px>=0&&px<n&&py>=0&&py<m&&(a[px][py]=='.'||a[px][py]=='D')&&!visi[px][py])        {            visi[px][py]=1;            dfs(px,py,step+1);            visi[px][py]=0;  //回溯        }    }    //return;}int main(){    int i,j;    while(scanf("%d%d%d",&n,&m,&sum))    {        if(n==0&&m==0&&sum==0)            break;        for(i=0;i<n;i++)            scanf("%s",a[i]);        flag=0;        for(i=0;i<n;i++)  //找起点        {            for(j=0;j<m;j++)            {               if(a[i][j]=='S')               {                  sta.x=i; sta.y=j;                  flag=1; break;               }            }            if(flag) break;        }        flag=0;        for(i=0;i<n;i++)  //找终点        {            for(j=0;j<m;j++)            {               if(a[i][j]=='D')               {                  en.x=i; en.y=j;                  flag=1; break;               }            }            if(flag) break;        }        flag=0;        memset(visi,0,sizeof(visi));        visi[sta.x][sta.y]=1;        dfs(sta.x,sta.y,0);        if(flag) puts("YES");        else puts("NO");    }    return 0;}//703MS 228K