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１、

２、题目：

Problem A

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, with B 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 Bfrom 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

Swedish National Contest

类似于跳棋，当两颗石子左或者右有空位置时，移动。每次转移之后移去经过的石子。

思路：

有12个格子，所以状态最多有2^12=4096个。把每次搜索过的状态存在dp[]数组中，以后再次查询类似的直接返回即可。

#include 
   
    #include 
    
     #include 
     
      #define min(a,b) (((a) < (b)) ? (a) : (b))int dp[4100];int solve(int n){    if (dp[n] != -1)        return dp[n];    dp[n] = 0;    for (int i = 0; i < 12; ++i)        if (n & (1 << i))            dp[n] += 1;    for (int i = 0; i < 10; ++i)    {        int t;        if ((n&(1<
      
       << i);            t &= ~(1 << (i+1));            t |= 1 << (i+2);            dp[n] = min(dp[n], solve(t));        }        if (!(n&(1<
       
        << (i+1));            t &= ~(1 << (i+2));            t |= 1 << i;            dp[n] = min(dp[n], solve(t));        }    }    return dp[n];}int main(){    memset(dp, -1, sizeof(dp));    int cases;    scanf("%d", &cases);    while (cases--)    {        char str[20];        int n = 0;        scanf("%s", str);        for (int i = 0; i < 12; ++i)            if (str[i] == 'o')                n ^= 1 << i;        printf("%d\n", solve(n));    }    return 0;}
       
      
     
    
   

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