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[USACO 6.5.4]The Clocks

2017-01-23 12:41 309 查看
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;
}
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