Hdu 1401 Solitaire 双向bfs-白红宇

Hdu 1401 Solitaire 双向bfs

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发布时间：2019-05-23

本文共 3508 字，大约阅读时间需要 11 分钟。

题目：

Solitaire is a game played on a chessboard 8x8. The rows and columns of the chessboard are numbered from 1 to 8, from the top to the bottom and from left to right respectively.

There are four identical pieces on the board. In one move it is allowed to:

> move a piece to an empty neighboring field (up, down, left or right),

> jump over one neighboring piece to an empty field (up, down, left or right).

There are 4 moves allowed for each piece in the configuration shown above. As an example let's consider a piece placed in the row 4, column 4. It can be moved one row up, two rows down, one column left or two columns right.

Write a program that:

> reads two chessboard configurations from the standard input,

> verifies whether the second one is reachable from the first one in at most 8 moves,

> writes the result to the standard output.

Input

Each of two input lines contains 8 integers a1, a2, ..., a8 separated by single spaces and describes one configuration of pieces on the chessboard. Integers a2j-1 and a2j (1 <= j <= 4) describe the position of one piece - the row number and the column number respectively. Process to the end of file.

Output

The output should contain one word for each test case - YES if a configuration described in the second input line is reachable from the configuration described in the first input line in at most 8 moves, or one word NO otherwise.

Sample Input

4 4 4 5 5 4 6 52 4 3 3 3 6 4 6

Sample Output

YES

思路：

第一次做的是用的单向bfs，然后简单的剪了一下枝，然后tle。

然后网上的都是用的双向bfs，于是就学了一手双向bfs，建了两个队列和两个hash表，然后从起点和终点开始搜索，双向bfs一次是搜索队列的一层，而不仅仅是一个节点。

当两个hash表中都有那个元素的时候就说明找到了。

hash表的建立是运用了移位运算，因为每一位是0-7，所以只占三位，因此表示每十进制一位数的时候只需要移动三位就可以了。

代码如下：

#include 
   
    #include 
    
     #include 
     
      #include 
      
       #include 
       #include 
        
         using namespace std;typedef long long ll;struct pos{ int x,y;};pos a[4],b[4];int loc[4][2]={ {1,0},{-1,0},{0,-1},{0,1}};map
         
          vis[2];struct node{ int step; pos a[4];};int compare(pos a,pos b){ return a.x!=b.x? a.x
          
           
            =0&&a.x<8&&a.y>=0&&a.y<8) return 1; return 0;}int isSame (pos a[],int x){ for (int i=0;i<4;i++) { if(i!=x&&a[i].x==a[x].x&&a[i].y==a[x].y) return 1; } return 0;}queue
            
             q[2];int bfs (int x){ node now,next; int Size=q[x].size(); while(Size--) { now=q[x].front(); q[x].pop(); ll t=change(now.a); for (int i=0;i<4;i++) { for (int j=0;j<4;j++) { next=now; next.step++; next.a[i].x+=loc[j][0]; next.a[i].y+=loc[j][1]; if(judge(next.a[i])) { if(isSame(next.a,i)) { next.a[i].x+=loc[j][0]; next.a[i].y+=loc[j][1]; if(judge(next.a[i])) { if(!isSame(next.a,i)) { ll t=change(next.a); if(vis[x^1][t]==1) { return 1; } else { vis[x][t]=1; q[x].push(next); } } } } else { ll t=change(next.a); if(vis[x^1][t]==1) { return 1; } else { vis[x][t]=1; q[x].push(next); } } } } } } return 0;}int tbfs (){ node now1,now2; for (int i=0;i<4;i++) { now1.a[i]=a[i]; now2.a[i]=b[i]; } q[0].push(now1); ll t1=change(now1.a); vis[0][t1]=1; q[1].push(now2); ll t2=change(now2.a); vis[1][t2]=1; if(t1==t2) return 1; int step=0; while(!q[0].empty()||!q[1].empty()) { if(step>=4) return 0; step++; if(bfs(0)) return 1; if(bfs(1)) return 1; } return 0;}int main(){ while(scanf("%d%d",&a[0].x,&a[0].y)!=EOF) { vis[0].clear(); vis[1].clear(); while(!q[0].empty()) q[0].pop(); while(!q[1].empty()) q[1].pop(); scanf("%d%d%d%d%d%d",&a[1].x,&a[1].y,&a[2].x,&a[2].y,&a[3].x,&a[3].y); for (int i=0;i<4;i++) a[i].x--,a[i].y--; for (int i=0;i<4;i++) { scanf("%d%d",&b[i].x,&b[i].y); b[i].x--,b[i].y--; } if(tbfs()) printf("YES\n"); else printf("NO\n"); } return 0;}

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