[USACO 6.5.4]The Clocks

题目大意

　　给出3*3共9个时钟的初始状态和结束状态(即指针全指向12)以及一系列的操作方法,求操作序列.详情请看题面.

题解

　　1.操作顺序是没有关系的.

　　2.每个操作只能执行0~3次.

代码

/*
TASK:clocks
LANG:C++
*/
#include <cstdio>
#include <cstring>

using namespace std;

const char mtd[9][7] = {"ABDE\0",
                        "ABC\0",
                        "BCEF\0",
                        "ADG\0",
                        "BDEFH\0",
                        "CFI\0",
                        "DEGH\0",
                        "GHI\0",
                        "EFHI\0"};
const int a[4] = {1, 10, 100, 1000};
const int b[4] = {0, 1, 11, 111};

int clock[9], rot[9], cnt[9], path[30], np, now;

void dfs(int dep, int dnow)
{
    if (dep == 9)
    {
        bool flag = true;
        for (int i = 0; i < 9; ++i)
            if (clock[i] != rot[i] % 4)
            {
                flag = false;
                break;
            }
        if (flag && dnow < now)
        {
            np = 0;
            for (int i = 0; i < 9; ++i)
                for (int j = 0; j < cnt[i]; ++j)
                    path[np++] = i + 1;    
            now = dnow;
        }
        return;
    }
    for (int i = 0; i < 4; ++i)
    {
        cnt[dep] = i;
        for (int lv = 0; lv < strlen(mtd[dep]); ++lv) rot[mtd[dep][lv] - 'A'] += i;
        dfs(dep + 1, dnow * a[i] + (dep+1) * b[i]);
        for (int lv = 0; lv < strlen(mtd[dep]); ++lv) rot[mtd[dep][lv] - 'A'] -= i;
    }
}

int main()
{
    freopen("clocks.in", "r", stdin);
    freopen("clocks.out", "w", stdout);
    for (int i = 0; i < 9; ++i)
    {
        scanf("%d", &clock[i]);
        clock[i] = (4 - clock[i]/3) % 4;
    }
    memset(rot, 0, sizeof(rot));
    now = 0x7fffffff;
    dfs(0, 0);
    for (int i = 0; i < np-1; ++i) printf("%d ", path[i]);
    printf("%d\n", path[np-1]);
    return 0;
}