poj 2965 枚举搜索

这题跟POJ1753 flip game 思想一样。。唯一不同得是这题要保存方案。

用c++一直超时，g++500多ms.

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <queue>
#include <stack>
using namespace std;

int visit[70000];
char mp[1100];

struct node
{
 int state;
 int num;
}p;

struct pnode
{
 int x, y;
}px[70000];

int path[70000];

int STATE;

int flip_num[16] = {63624,62532,61986,61713,36744,20292,12066,7953,35064,
               17652,8946,4593,34959,17487, 8751, 4383};
void print( int x )
{
  
  stack<int>qx;
  while( x != STATE ) 
  {
    qx.push(x);
    x = path[x];
    //printf("%d %d\n", px[x].x + 1, px[x].y + 1);
  }
  while( !qx.empty( ) )
  {
    int t = qx.top();
    qx.pop( );
    printf("%d %d\n", px[t].x + 1, px[t].y + 1);
  }

}

void print1( int x )
{
   if( x != STATE )
   {
       print1( path[x] );
       printf("%d %d\n", px[x].x + 1, px[x].y + 1);
   }


}

void bfs( )
{
  int flag = 0;
  node pp;
  queue<node>q;
  q.push(p);
  visit[p.state] = 1;
  while( !q.empty())
  {
     p = q.front();
     q.pop();
     pp.state = p.state;
     for( int i = 0; i < 16; i++)
     {
         p.state = pp.state;
         //p.state = p.state ^ ( flip_num[15-i] ); 500多ms..
         int row = i / 4;
         int col = i - row * 4;
         
         p.state = p.state ^ ( 1 << i );
         
         for( int j = 0; j < 4; j++)
         {
            p.state = p.state ^ (1 << (row * 4 + j ) ); 
            p.state = p.state ^ (1 << (4 * j + col ) );
         }
         
         if( p.state == 0 )
         {
          cout<<p.num+1<<endl;
          px[p.state].x = row;
          px[p.state].y = col;
          path[p.state] = pp.state;
          print1( 0 );
          return;
         }
         if( visit[p.state] == 0 )
         {
            px[p.state].x = row;
            px[p.state].y = col;
            path[p.state] = pp.state;
            node temp;
            temp.state = p.state;
            temp.num = p.num + 1;
            visit[p.state] = 1;
            q.push(temp);
         }

     }

  }
  
}

int main( )
{
  p.state = 0;
  p.num = 0;
  memset(visit, 0, sizeof(visit));
  for(int i = 0; i < 16; i++)
  {
    cin>>mp[i];
    if( mp[i] == '+')
      p.state += (1<<i);
  }
  STATE = p.state;
  bfs( );
  return 0;
}