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;

}