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  • UVa 10651 Pebble Solitaire 状态压缩 dp

    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 AB, 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

    5

    ---oo-------

    -o--o-oo----

    -o----ooo---

    oooooooooooo

    oooooooooo-o

    1

    2

    3

    12

    1

     



    ----------------------------------------------------------

    啊,难得把状态压对一次。

    令黑色为1,白色为0。

    f[x]表示状态x所能完成的最大转换数。f[x]=max( f[t]+1 ) (x可以到达t)

    黑色个数-f[x]即为答案。

    -----------------------------------------------------------

    #include <iostream>
    #include <cstdio>
    #include <cstring>
    
    using namespace std;
    
    //110=6 001=1 011=3 100=4
    
    int f[1<<12];
    int tmp;
    
    int dfs(int x)
    {
        //cerr<<"x= "<<x<<endl;
        //getchar();
        if (f[x]!=-1) return f[x];
        int ret=0;
        for (int i=0;i<10;i++)
        {
            int t=x;
            if ( (x&(1<<(i+2))) && (x&(1<<(i+1))) && (x&(1<<i))==0 )
            {
                t^=(6<<i);
                t|=(1<<i);
                //cerr<<"check ok x= "<<x<<" i= "<<i<<" 6<<i= "<<(6<<i)<<" t="<<t<<endl;
                ret=max( ret, dfs(t)+1 );
            }
            t=x;
            if ( (x&(1<<(i+2)))==0 && (x&(1<<(i+1))) && (x&(1<<i)) )
            {
                t^=(3<<i);
                t|=(4<<i);
                //cerr<<"check ok x= "<<x<<" t="<<t<<endl;
                ret=max( ret, dfs(t)+1 );
            }
        }
        f[x]=ret;
        return ret;
    }
    
    int main()
    {
        int T;
        char s[20];
        int black,bit;
        memset(f,-1,sizeof(f));
        scanf("%d",&T);
        while (T--)
        {
            bit=0;
            black=0;
            scanf("%s",s);
            for (int i=0;s[i];i++)
            {
                if (s[i]=='o')
                {
                    bit|=(1<<i);
                    black++;
                }
            }
            //读入完毕
            int ans=dfs(bit);
            printf("%d\n",black-ans);
        }
        return 0;
    }
    




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  • 原文地址:https://www.cnblogs.com/cyendra/p/3226396.html
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