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
|
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啊,难得把状态压对一次。
令黑色为1,白色为0。
f[x]表示状态x所能完成的最大转换数。f[x]=max( f[t]+1 ) (x可以到达t)
黑色个数-f[x]即为答案。
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#include <iostream> #include <cstdio> #include <cstring> using namespace std; //110=6 001=1 011=3 100=4 int f[1<<12]; int tmp; int dfs(int x) { //cerr<<"x= "<<x<<endl; //getchar(); if (f[x]!=-1) return f[x]; int ret=0; for (int i=0;i<10;i++) { int t=x; if ( (x&(1<<(i+2))) && (x&(1<<(i+1))) && (x&(1<<i))==0 ) { t^=(6<<i); t|=(1<<i); //cerr<<"check ok x= "<<x<<" i= "<<i<<" 6<<i= "<<(6<<i)<<" t="<<t<<endl; ret=max( ret, dfs(t)+1 ); } t=x; if ( (x&(1<<(i+2)))==0 && (x&(1<<(i+1))) && (x&(1<<i)) ) { t^=(3<<i); t|=(4<<i); //cerr<<"check ok x= "<<x<<" t="<<t<<endl; ret=max( ret, dfs(t)+1 ); } } f[x]=ret; return ret; } int main() { int T; char s[20]; int black,bit; memset(f,-1,sizeof(f)); scanf("%d",&T); while (T--) { bit=0; black=0; scanf("%s",s); for (int i=0;s[i];i++) { if (s[i]=='o') { bit|=(1<<i); black++; } } //读入完毕 int ans=dfs(bit); printf("%d\n",black-ans); } return 0; }