Pebble Solitaire（UVA 10651）

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, whereA 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 B from 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

  

题解：dfs，由于有12个位置，需要2^12个数来记录中间值

代码：c++ 

//Memory:0Kb ; Time:12ms

code1:使用数组记录中间结果

#include<stdio.h>#define min(x,y) (x<y?x:y)char buf[13];int F[4100];int dfs(int s){    if(F[s]<13) return F[s];  //avoid the repeat calculator    int Min=12;    for(int a=0;a<10;a++){        if(((s>>a)&7)==6||((s>>a)&7)==3){            Min=min(Min,dfs(s^(7<<a)));        }    }    if(Min==12){        for(int i=0;i<12;i++){            Min-=!(s&(1<<i));        }    }    return F[s]=Min;}int main(){    int n;    scanf("%d",&n);    while(n--){        scanf("%s",buf);        int s;        s=0;        for(int a=0;a<12;a++){            s<<=1;            if(buf[a]=='o'){                s+=1;            }        }        for(int a=0;a<4100;a++){            F[a]=13;        }        printf("%d\n",dfs(s));    }    return 0;}

code2：

#include<stdio.h>#define min(x,y) (x<y?x:y)char buf[13];int dfs(int s){    int Min=12;    for(int a=0;a<10;a++){        if(((s>>a)&7)==6||((s>>a)&7)==3){            Min=min(Min,dfs(s^(7<<a)));        }    }    if(Min==12){        for(int i=0;i<12;i++){            Min-=!(s&(1<<i));        }    }    return Min;}int main(){    int n;    scanf("%d",&n);    while(n--){        scanf("%s",buf);        int s;        s=0;        for(int a=0;a<12;a++){            s<<=1;            if(buf[a]=='o'){                s+=1;            }        }        printf("%d\n",dfs(s));    }    return 0;}