Pebble Solitaire

 

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 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

题意  给你一个长度为12的字符串  由字符‘-‘和字符‘o‘组成  当中"-oo"和"oo-"分别能够通过一次转换变为"o--"和"--o"  能够发现每次转换o都少了一个  仅仅需求出给你的字符串做多能转换多少次即可了。

令d[s]表示字符串s最多能够转换的次数  若s能够通过一次转换变为字符串t  有d[s]=max(d[s],d[t]+1);

#include<iostream>
#include<string>
#include<map>
using namespace std;
map<string, int> d;
int n, ans;
string t, S;

int dp (string s)
{
    if (d[s] > 0) return d[s];
    d[s] = 1;
    for (int i = 0; i < 10; ++i)
    {
        if (s[i] == ‘o‘ && s[i + 1] == ‘o‘ && s[i + 2] == ‘-‘)
        {
            t = s;
            t[i] = t[i + 1] = ‘-‘;
            t[i + 2] = ‘o‘;
            d[s] = max (d[s], dp (t) + 1);
        }
        if (s[i] == ‘-‘ && s[i + 1] == ‘o‘ && s[i + 2] == ‘o‘)
        {
            t = s;
            t[i] = ‘o‘;
            t[i + 1] = t[i + 2] = ‘-‘;
            d[s] = max (d[s], dp (t) + 1);
        }
    }
    return d[s];
}

int main()
{
    cin >> n;
    while (n--)
    {
        ans = 1;
        cin >> S;
        for (int i = 0; i < 12; ++i)
            if (S[i] == ‘o‘) ans++;
        ans -= dp (S);
        cout << ans << endl;
    }
    return 0;
}