hdu1010搜尋+剪枝

Tempter of the Bone

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 42078    Accepted Submission(s): 11382


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

Sample Output

NO YES 

這題要解釋一下，剪枝剪出翔了。奇偶剪枝以前都沒聽說過。反正要各種注意，

DFS用回溯法架構來做#include <iostream>
#include <cmath>
using namespace std;
int N,M,T;

int time_used;

char map[7][7];
int dirx[4]={-1,0,1,0};
int diry[4]={0,1,0,-1};
int sx,sy;
int ex,ey;
bool flag;
int Count;

void DFS(int x,int y)
{
     int xx,yy;
     if(flag)
         return;
     if((x == ex && y == ey)&& (time_used == T))    //達到條件退出 
     {
         flag = true;
         return;
     }
     if((T-time_used)%2 != (abs(x-ex)+abs(y-ey))%2)  //奇偶減枝法
         return;
     if(abs(x-ex)+abs(y-ey) > T-time_used)                             //當前點到終點的最短時間若比剩餘的時間還長的話
         return;
     

     
     for(int i = 0; i < 4;i++)      //向四個方向擴充 
     {
         xx = x + dirx[i];
         yy = y + diry[i];
         if( (xx >= 0) && (xx < N) && (yy >= 0) && (yy <M) && map[xx][yy] != 'X')
         {
                  
                  map[x][y] = 'X';
                  time_used++;
                  DFS(xx,yy);
                  time_used--;       
                  map[x][y] = '.';
                  
         }
     }
}

int main()
{
    
    while(cin>>N>>M>>T)
    {
       if(!N && !M && !T)
             break;
       Count = 0;
       for(int i = 0;i < N;i++)
       {
            for(int j = 0;j < M;j++)
            {
                cin>>map[i][j];
                if(map[i][j] == 'S')
                {
                    sx = i;
                    sy = j;
                }
                if(map[i][j] == 'D')
                {
                    ex = i;
                    ey = j;
                }
                else if(map[i][j] == '.')
                {
                     Count++;
                }
            }
            
       }
       if(Count +1 < T)      // 如果總共可以走的步數比給定時間少 
       {
           cout<<"NO"<<endl;
           continue;
       }
       time_used = 0;
       flag = false;
       DFS(sx,sy);
       if(flag == true)
           cout<<"YES"<<endl;
       else
           cout<<"NO"<<endl;
       
    }
    return 0;
}

擴充：奇偶剪枝　　是資料結構的搜尋中，剪枝的一種特殊小技巧。 　　現假設起點為(sx,sy)，終點為（ex,ey），給定t步恰好走到終點， 　　

s
|
|
|
+	—	—	—	e

（“|”豎走，“—”橫走，“+”轉彎），易證abs(ex-sx)+abs(ey-sy)為此問題類中任意情況下，起點到終點的最短步數，記做step，此處step1=8； 　　

s	—	—	—
	—	—	+
|	+
|
+	—	—	—	e

，為一般情況下非最短路徑的任意走法舉例，step2=14； 　　step2-step1=6，位移路徑為6，偶數（易證）； 　　故，若t-[abs(ex-sx)+abs(ey-sy)]結果為非偶數（奇數），則無法在t步恰好到達； 　　返回，false；  　　反之亦反。