uva 10651 - Pebble Solitaire

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

 

这道题是一道简单状压dp，只要熟练掌握位运算即可，实际上这道题只要判断是否存在‘-oo'和’oo-'的情况，然后对这三位按位取反得到一个新的状态，一直到状态没有上述两种情况存在，就统计二进制1的位数。其余过程就是简单的记忆化搜索。

代码：

#include<cstdio>#include<cstring>#include<iostream>using namespace std;char ch;const int Maxn=1<<12;int dp[Maxn];bool check(int x,int i){    if(x&1<<i-1&&x&1<<i&&!(x&1<<i+1)) return true;    if(!(x&1<<i-1)&&x&1<<i&&x&1<<i+1) return true;    return false;}int countbit(int x){    int ans=0;    while(x){        ans+=x&1;        x>>=1;    }    return ans;}int dfs(int x){    int &ans=dp[x];    if(ans!=13) return ans;    bool flag=true;    for(int i=1;i<11;i++)        if(check(x,i)){            ans=min(ans,dfs(x^1<<i-1^1<<i^1<<i+1));            flag=false;        }    if(flag) return countbit(x);    return ans;}int main(){    int n,num;    scanf("%d%*c",&n);    while(n--){        num=0;        for(int i=0;i<12;i++){            scanf("%c",&ch);            if(ch=='o') num|=1<<i;        }        scanf("%*c");        for(int i=0;i<Maxn;i++) dp[i]=13;        printf("%d\n",dfs(num));    }return 0;}