Pebble
Solitaire
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 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
|
题意 给你一个长度为12的字符串 由字符'-'和字符'o'组成 当中"-oo"和"oo-"分别能够通过一次转换变为"o--"和"--o" 能够发现每次转换o都少了一个 仅仅需求出给你的字符串做多能转换多少次即可了。
令d[s]表示字符串s最多能够转换的次数 若s能够通过一次转换变为字符串t 有d[s]=max(d[s],d[t]+1);
#include<iostream>
#include<string>
#include<map>
using namespace std;
map<string, int> d;
int n, ans;
string t, S; int dp (string s)
{
if (d[s] > 0) return d[s];
d[s] = 1;
for (int i = 0; i < 10; ++i)
{
if (s[i] == 'o' && s[i + 1] == 'o' && s[i + 2] == '-')
{
t = s;
t[i] = t[i + 1] = '-';
t[i + 2] = 'o';
d[s] = max (d[s], dp (t) + 1);
}
if (s[i] == '-' && s[i + 1] == 'o' && s[i + 2] == 'o')
{
t = s;
t[i] = 'o';
t[i + 1] = t[i + 2] = '-';
d[s] = max (d[s], dp (t) + 1);
}
}
return d[s];
} int main()
{
cin >> n;
while (n--)
{
ans = 1;
cin >> S;
for (int i = 0; i < 12; ++i)
if (S[i] == 'o') ans++;
ans -= dp (S);
cout << ans << endl;
}
return 0;
}