uva 10651- Pebble Solitaire(状态压缩ＤＰ)待看。。。

１、http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1592

２、题目：

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

Swedish National Contest

类似于跳棋，当两颗石子左或者右有空位置时，移动。每次转移之后移去经过的石子。

思路：

有12个格子，所以状态最多有2^12=4096个。把每次搜索过的状态存在dp[]数组中，以后再次查询类似的直接返回即可。

３、AC代码：

#include <cstdio>
#include <cstdlib>
#include <cstring>

#define min(a,b) (((a) < (b)) ? (a) : (b))

int dp[4100];

int solve(int n)
{
if (dp
!= -1)
return dp
;

dp
= 0;
for (int i = 0; i < 12; ++i)
if (n & (1 << i))
dp
+= 1;

for (int i = 0; i < 10; ++i)
{
int t;
if ((n&(1<<i)) && (n&(1<<(i+1))) && !(n&(1<<(i+2))))
{
t = n;
t &= ~(1 << i);
t &= ~(1 << (i+1));
t |= 1 << (i+2);
dp
= min(dp
, solve(t));
}

if (!(n&(1<<i)) && (n&(1<<(i+1))) && (n&(1<<(i+2))))
{
t = n;
t &= ~(1 << (i+1));
t &= ~(1 << (i+2));
t |= 1 << i;
dp
= min(dp
, solve(t));
}
}
return dp
;
}

int main()
{
memset(dp, -1, sizeof(dp));

int cases;
scanf("%d", &cases);
while (cases--)
{
char str[20];
int n = 0;

scanf("%s", str);
for (int i = 0; i < 12; ++i)
if (str[i] == 'o')
n ^= 1 << i;

printf("%d\n", solve(n));
}
return 0;
}