Pebble Solitaire - UVa 10651 状压dp+记忆化搜素

Problem A
Pebble Solitaire
Input: standard input
Output: standard output
Time Limit: 1 second
 

Pebble solitaire is an interesting game. This is a game where you are given a board with an arrangement of small cavities, initially all but one occupied by a pebble each. The aim of the game is to remove as many pebbles as possible from the board. Pebbles disappear from the board as a result of a move. A move is possible if there is a straight line of three adjacent cavities, let us call them A, B, and C, withB in the middle, where A is vacant, but B and C each contain a pebble. The move constitutes of moving the pebble from C to A, and removing the pebble in B from the board. You may continue to make moves until no more moves are possible.

 

In this problem, we look at a simple variant of this game, namely a board with twelve cavities located along a line. In the beginning of each game, some of the cavities are occupied by pebbles. Your mission is to find a sequence of moves such that as few pebbles as possible are left on the board.

 

Input

The input begins with a positive integer n on a line of its own. Thereafter n different games follow. Each game consists of one line of input with exactly twelve characters, describing the twelve cavities of the board in order. Each character is either '-' or 'o' (The fifteenth character of English alphabet in lowercase). A '-' (minus) character denotes an empty cavity, whereas a 'o' character denotes a cavity with a pebble in it. As you will find in the sample that there may be inputs where no moves is possible.

 

Output

For each of the n games in the input, output the minimum number of pebbles left on the board possible to obtain as a result of moves, on a row of its own.

 

Sample Input                              Output for Sample Input

5

---oo-------

-o--o-oo----

-o----ooo---

oooooooooooo

oooooooooo-o

1

2

3

12

1

 

题意：相邻的两块石头，可以让一块跳过另一块，然后把被跳过的拿走，问最后最少剩多少石头。

思路：01的二进制表示状况，然后枚举所有可能的情况往下走，然后记录已经访问过的情况，减少复杂度。

AC代码如下：

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int num[5010],vis[5010];void dfs(int S){ if(vis[S]==1)   return;  vis[S]=1;  int k=0,p,i;  p=S;  while(p)  { if(p&1)     k++;    p/=2;  }  for(i=0;i<=9;i++)   if( ((3<<i)&S)==(3<<i) && ( (1<<(i+2))&S)==0)   { p=S-(3<<i)+ (1<<(i+2));     dfs(p);     k=min(k,num[p]);   }  for(i=1;i<=10;i++)   if( ((3<<i)&S)==(3<<i) && ( (1<<(i-1))&S)==0)   { p=S-(3<<i)+ (1<<(i-1));     dfs(p);     k=min(k,num[p]);   }  num[S]=k;}char s[20];int main(){ int t,i,j,k,S;  scanf("%d",&t);  while(t--)  { scanf("%s",s);    S=0;    for(i=0;i<=11;i++)    { S*=2;      if(s[i]=='o')       S+=1;    }    dfs(S);    printf("%d\n",num[S]);  }}