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 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

5

---oo-------

-o--o-oo----

-o----ooo---

oooooooooooo

oooooooooo-o

1

2

3

12

1

 

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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。

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#include <iostream>#include <cstdio>#include <cstring>using namespace std;//110=6 001=1 011=3 100=4int 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;}