Uva--10651（动规，记忆化搜索）

2014-08-02 19:42:38

 

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, with B in the middle, whereA 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

思路：也是一种记忆化搜索，对于每层递归，扫整个字符串，看看哪里可以进行变幻，能变幻就继续递归进去。

/*************************************************************************
    > File Name: l.cpp
    > Author: Nature
    > Mail: 564374850@qq.com
    > Created Time: Thu 31 Jul 2014 05:30:20 PM CST
************************************************************************/

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <algorithm>
using namespace std;

int n;
char s[15];

int Count(int p){
    int cnt = 0;
    while(p){
        cnt += p & 1;
        p >>= 1;
    }
    return cnt;
}

int Judge(int p,int i){
    if(!(p & (1 << i)) && (p & (1 << (i + 1))) && (p & (1 << (i + 2))))
        return 1;
    else if((p & (1 << i)) && (p & (1 << (i + 1))) && !(p & (1 << (i + 2))))
        return 2;
    return 0;
}

int Solve(int p){
    int cnt = Count(p);
    int tmin = 1e9;
    int flag;
    int t;
    for(int i = 0; i < 10; ++i){
        flag = Judge(p,i);
        if(flag == 1){
            t = p;
            t |= (1 << i);
            t ^= (1 << (i + 1));
            t ^= (1 << (i + 2));
            //printf("flag 1 , t : %d\n",t);
            tmin = min(tmin,Solve(t));
        }
        else if(flag == 2){
            t = p;
            t ^= (1 << i);
            t ^= (1 << (i + 1));
            t |= (1 << (i + 2));
            //printf("flag 2 , t : %d\n",t);
            tmin = min(tmin,Solve(t));
        }
    }
    return min(tmin,cnt);
}

int main(){
    int num;
    scanf("%d",&n);
    while(n--){
        scanf("%s",s);
        num = 0;
        for(int i = 0; i < 12; ++i){
            if(s[i] == 'o')
                num |= (1 << i);
        }
        //printf("num : %d\n",num);
        printf("%d\n",Solve(num));
    }
    return 0;
}