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.
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.
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 其中"-oo"和"oo-"分别可以通过一次转换变为"o--"和"--o" 可以发现每次转换o都少了一个 只需求出给你的字符串做多能转换多少次就行了;
令d[s]表示字符串s最多可以转换的次数 若s可以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; }
UVa 10651 Pebble Solitaire(DP 记忆化搜索)
原文地址:http://blog.csdn.net/iooden/article/details/38070777