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[n] != -1)return dp[n];dp[n] = 0;for (int i = 0; i < 12; ++i)if (n & (1 << i))dp[n] += 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[n] = min(dp[n], 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[n] = min(dp[n], solve(t));}}return dp[n];
}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;
}