POJ 1166 The clocks

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bobten2008

最新推荐文章于 2021-02-03 20:16:45 发布

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分类专栏： BFS 位运算文章标签： numbers output each struct input c

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本文介绍了一个经典的九宫格时钟谜题，并通过位运算结合BFS算法求解。该问题要求将一个3x3排列的九个时钟全部调整到12点方向，提供了九种旋转方式。

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The Clocks

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 8928		Accepted: 3479

Description

|-------|    |-------|    |-------|

|       |    |       |    |   |   |

|---O   |    |---O   |    |   O   |

|       |    |       |    |       |

|-------|    |-------|    |-------|

    A            B            C    



|-------|    |-------|    |-------|

|       |    |       |    |       |

|   O   |    |   O   |    |   O   |

|   |   |    |   |   |    |   |   |

|-------|    |-------|    |-------|

    D            E            F    



|-------|    |-------|    |-------|

|       |    |       |    |       |

|   O   |    |   O---|    |   O   |

|   |   |    |       |    |   |   |

|-------|    |-------|    |-------|

    G            H            I    

(Figure 1)

There are nine clocks in a 3*3 array (figure 1). The goal is to return all the dials to 12 o'clock with as few moves as possible. There are nine different allowed ways to turn the dials on the clocks. Each such way is called a move. Select for each move a number 1 to 9. That number will turn the dials 90' (degrees) clockwise on those clocks which are affected according to figure 2 below.

Move   Affected clocks

 

 1         ABDE

 2         ABC

 3         BCEF

 4         ADG

 5         BDEFH

 6         CFI

 7         DEGH

 8         GHI

 9         EFHI    

   (Figure 2)

Input

Your program is to read from standard input. Nine numbers give the start positions of the dials. 0=12 o'clock, 1=3 o'clock, 2=6 o'clock, 3=9 o'clock.

Output

Your program is to write to standard output. Output a shortest sorted sequence of moves (numbers), which returns all the dials to 12 o'clock. You are convinced that the answer is unique.

Sample Input

3 3 0
2 2 2
2 1 2

Sample Output

4 5 8 9

Source

IOI 1994

这题用位运算+BFS过的，效率不是很高，但是毕竟过了。网上利用数论的mod和矩阵运算有更快的方法，这个留着以后有时间再研究了

思路比较简单，关键是位运算,由于一共有9个clock，每个clock最多4种状态因此每个clock可以用2bit来表示，9个clock一共需要

18bit来表示，0-1位表示第一个clock的状态,2-3位表示第二个clock的状态，一次类推，这样一共需要2 ^ 18 - 1 = 262143种状态

#include <iostream> #include <queue> #include <cmath> #define MAX_N 262150 using namespace std; bool v[MAX_N + 1]; int state; int expv[20]; int move[10][6] = { {0, 0, 0, 0, 0, 0}, {4, 1, 2, 4, 5, 0}, {3, 1, 2, 3, 0, 0}, {4, 2, 3, 5, 6, 0}, {3, 1, 4, 7, 0, 0}, {5, 2, 4, 5, 6, 8}, {3, 3, 6, 9, 0, 0}, {4, 4, 5, 7, 8, 0}, {3, 7, 8, 9, 0, 0}, {4, 5, 6, 8, 9, 0}}; struct node { int curState; int preMove; int preState; node() { curState = preMove = 0; } }nodes[MAX_N + 1]; //位运算操作，从state状态移动第which个clock void moveClock(int &state, int which) { int bit1, bit2; if(state & expv[2 * (which - 1)]) bit1 = 1; else bit1 = 0; if(state & expv[2 * (which - 1) + 1]) bit2 = 1; else bit2 = 0; if(bit1 == 0) state = state + expv[2 * (which - 1)]; else { if(bit2 == 0) { state = state - expv[2 * (which - 1)]; state = state + expv[2 * (which - 1) + 1]; } else { state = state - expv[2 * (which - 1)]; state = state - expv[2 * (which - 1) + 1]; } } } void printRes(int curState) { if(curState == state) return; else { printRes(nodes[curState].preState); cout<<nodes[curState].preMove<<" "; } } void bfs() { queue<node *> bfsq; bfsq.push(&nodes[state]); v[state] = true; int curState, newState; while(!bfsq.empty()) { node * curNode = bfsq.front(); bfsq.pop(); curState = curNode->curState; if(curState == 0) { printRes(curState); cout<<endl; return; } for(int i = 1; i <= 9; i++) { newState = curState; for(int j = 1; j <= move[i][0]; j++) { moveClock(newState, move[i][j]); } if(!v[newState]) { v[newState] = true; nodes[newState].preMove = i; nodes[newState].preState = curState; bfsq.push(&nodes[newState]); } } } } int main() { int i, temp; state = 0; for(i = 1; i <= 9; i++) { cin>>temp; state = state + temp * pow(double(4), i - 1); } expv[0] = 1; for(i = 1; i < 19; i++) expv[i] = expv[i - 1] * 2; for( i = 0; i <= MAX_N; i++) { nodes[i].curState = i; nodes[i].preMove = -1; } bfs(); return 0; }