usaco6.5.4 The Clocks_prob:clocks lang:pascal id:asiapea1-优快云博客

本文探讨了一道经典的算法题目——九个时钟的表盘需要通过特定的操作全部调整到12点方向。利用广度优先搜索(BFS)算法，找到最短的操作序列。文章附带了完整的C++代码实现，并通过了一系列测试案例验证了算法的有效性和正确性。

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一原题

The Clocks
IOI'94 - Day 2Consider nine clocks arranged in a 3x3 array thusly:

|-------|    |-------|    |-------|    
|       |    |       |    |   |   |    
|---O   |    |---O   |    |   O   |          
|       |    |       |    |       |           
|-------|    |-------|    |-------|    
    A            B            C

|-------|    |-------|    |-------|
|       |    |       |    |       |
|   O   |    |   O   |    |   O   |
|   |   |    |   |   |    |   |   |
|-------|    |-------|    |-------|
    D            E            F

|-------|    |-------|    |-------|
|       |    |       |    |       |
|   O   |    |   O---|    |   O   |
|   |   |    |       |    |   |   |
|-------|    |-------|    |-------|
    G            H            I

The goal is to find a minimal sequence of moves to return all the dials to 12 o'clock. Nine different ways to turn the dials on the clocks are supplied via a table below; each way is called a move. Select for each move a number 1 through 9 which will cause the dials of the affected clocks (see next table) to be turned 90 degrees clockwise.

Move	Affected clocks
1	ABDE
2	ABC
3	BCEF
4	ADG
5	BDEFH
6	CFI
7	DEGH
8	GHI
9	EFHI

Example

Each number represents a time according to following table:

9 9 12       9 12 12       9 12 12        12 12 12      12 12 12 
6 6 6  5 ->  9  9  9  8->  9  9  9  4 ->  12  9  9  9-> 12 12 12 
6 3 6        6  6  6       9  9  9        12  9  9      12 12 12

[But this might or might not be the `correct' answer; see below.]

PROGRAM NAME: clocks

INPUT FORMAT

Lines 1-3:

Three lines of three space-separated numbers; each number represents the start time of one clock, 3, 6, 9, or 12. The ordering of the numbers corresponds to the first example above.

SAMPLE INPUT (file clocks.in)

9 9 12
6 6 6
6 3 6

OUTPUT FORMAT

A single line that contains a space separated list of the shortest sequence of moves (designated by numbers) which returns all the clocks to 12:00. If there is more than one solution, print the one which gives the lowest number when the moves are concatenated (e.g., 5 2 4 6 < 9 3 1 1).

SAMPLE OUTPUT (file clocks.out)

4 5 8 9

二分析

没啥好说的。。BFS

三代码

运行结果：

USER: Qi Shen [maxkibb3]
TASK: clocks
LANG: C++

Compiling...
Compile: OK

Executing...
   Test 1: TEST OK [0.000 secs, 5280 KB]
   Test 2: TEST OK [0.000 secs, 5280 KB]
   Test 3: TEST OK [0.014 secs, 6072 KB]
   Test 4: TEST OK [0.042 secs, 6600 KB]
   Test 5: TEST OK [0.056 secs, 6864 KB]
   Test 6: TEST OK [0.196 secs, 7088 KB]
   Test 7: TEST OK [0.238 secs, 7392 KB]
   Test 8: TEST OK [0.224 secs, 5780 KB]
   Test 9: TEST OK [0.224 secs, 5812 KB]

All tests OK.

YOUR PROGRAM ('clocks') WORKED FIRST TIME! That's fantastic -- and a rare thing. Please accept these special automated congratulations.

AC代码：

/*
ID:maxkibb3
LANG:C++
PROB:clocks
*/

#include <cstdio>
#include <vector>
#include <queue>

using namespace std;

// #define local

const short action[9][9] = {
    {1, 1, 0, 1, 1, 0, 0, 0, 0},
    {1, 1, 1, 0, 0, 0, 0, 0, 0},
    {0, 1, 1, 0, 1, 1, 0, 0, 0},
    {1, 0, 0, 1, 0, 0, 1, 0, 0},
    {0, 1, 0, 1, 1, 1, 0, 1, 0},
    {0, 0, 1, 0, 0, 1, 0, 0, 1},
    {0, 0, 0, 1, 1, 0, 1, 1, 0},
    {0, 0, 0, 0, 0, 0, 1, 1, 1},
    {0, 0, 0, 0, 1, 1, 0, 1, 1}
};

bool vis[1000000];

struct Clock {
    short t[9];
    vector<short> trace;
}st;

inline int h(const Clock &c) {
    int ret = 0, base = 1;
    for(int i = 0; i < 9; i++) {
        ret += c.t[i] * base;
        base *= 4;
    }
    return ret;
}

int main() {
#ifndef local
    freopen("clocks.in", "r", stdin);
    freopen("clocks.out", "w", stdout);
#endif
    for(int i = 0; i < 9; i++) {
        scanf("%hd", &st.t[i]);
        st.t[i] = (st.t[i] / 3) % 4;
    }
    if(h(st) == 0) return 0;

    queue<Clock> q;
    vector<short> ans;
    q.push(st);
    vis[h(st)] = true;
    bool end = false;
    while(!q.empty()) {
        Clock hd = q.front(), nxt;
        q.pop();
        for(int i = 0; i < 9; i++) {
            if(end) break;
            for(int j = 0; j < 9; j++) {
                nxt.t[j] = (hd.t[j] + action[i][j]) % 4;
            }
            int hashval = h(nxt);
            if(vis[hashval]) continue;
            nxt.trace = hd.trace;
            nxt.trace.push_back(i);
            if(hashval == 0) {
                end = true;
                ans = nxt.trace;
            }
            q.push(nxt);
            vis[hashval] = true;
        }
        if(end) break;
    }

    printf("%d", ans[0] + 1);
    for(int i = 1; i < ans.size(); i++) {
        printf(" %d", ans[i] + 1);
    }
    printf("\n");

    return 0;
}