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
|
Swedish National Contest
#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <vector>
using namespace std;
struct node{
int a,b,c;
node(int a,int b,int c):a(a),b(b),c(c){}
};
string st;
int dp[1<<13];
int dfs(int mask){
if(dp[mask] !=-1) return dp[mask];
vector<int> v;
vector<node>pos;
v.clear();
pos.clear();
int tmp = mask;
for(int i = 0; i < 12; i++){
v.push_back(tmp%2);
tmp /= 2;
}
int i = 1,ret = v[0];
bool flag = 0;
while(i < 12){
if(v[i]==1){
ret++;
if((v[i+1]==1&&v[i-1]==0&&i+1<12)||(v[i+1]==0&&v[i-1]==1&&i+1<12)){
flag = 1;
pos.push_back(node(i-1,i,i+1));
}
}
i++;
}
if(!flag)
return dp[mask] = ret;
int ans = 1e8;
for(int i = 0; i < pos.size(); i++){
int tma = mask;
if(v[pos[i].a]==0){
tma ^= (1<<pos[i].b);
tma ^= (1<<pos[i].c);
tma |= (1<<pos[i].a);
}else{
tma ^= (1<<pos[i].b);
tma |= (1<<pos[i].c);
tma ^= (1<<pos[i].a);
}
ans = min(ans,dfs(tma));
}
return dp[mask] = ans;
}
int main(){
int ncase;
cin >> ncase;
memset(dp,-1,sizeof dp);
while(ncase--){
cin >> st;
int ms = 0;
for(int i = 0; i < st.size(); i++)
if(st[i]=='o') ms |= (1<<i);
cout<<dfs(ms)<<endl;
}
return 0;
}