USACO 3.2 Magic Squares

Magic Squares
IOI'96

Following the success of the magic cube, Mr. Rubik invented its planar version, called magic squares. This is a sheet composed of 8 equal-sized squares:

1	2	3	4
8	7	6	5

In this task we consider the version where each square has a different color. Colors are denoted by the first 8 positive integers. A sheet configuration is given by the sequence of colors obtained by reading the colors of the squares starting at the upper left corner and going in clockwise direction. For instance, the configuration of Figure 3 is given by the sequence (1,2,3,4,5,6,7,8). This configuration is the initial configuration.

Three basic transformations, identified by the letters `A', `B' and `C', can be applied to a sheet:

• 'A': exchange the top and bottom row,
• 'B': single right circular shifting of the rectangle,
• 'C': single clockwise rotation of the middle four squares.

Below is a demonstration of applying the transformations to the initial squares given above:

A:

8	7	6	5
1	2	3	4

B:

4	1	2	3
5	8	7	6

C:

1	7	2	4
8	6	3	5

All possible configurations are available using the three basic transformations.

You are to write a program that computes a minimal sequence of basic transformations that transforms the initial configuration above to a specific target configuration.

PROGRAM NAME: msquare

INPUT FORMAT

A single line with eight space-separated integers (a permutation of (1..8)) that are the target configuration.

SAMPLE INPUT (file msquare.in)

2 6 8 4 5 7 3 1

OUTPUT FORMAT

Line 1:	A single integer that is the length of the shortest transformation sequence.
Line 2:	The lexically earliest string of transformations expressed as a string of characters, 60 per line except possibly the last line.

SAMPLE OUTPUT (file msquare.out)

7
BCABCCB

——————————————————

就喜欢usaco题面一复制就下来了

8！很容易想到的是搜索，我们需要对一个长为8的全排列数列进行判重和记录

所以就是引入一个“中介数”

好吧，继续那个运用好几次的例子了……来源是一本余姚的书……

通过中介数(221)↑构造排列的过程如下：

1. 首先这个排列的四个数的位置都是未知的：_ _ _ _
2. 从左向右看中介数，第一个2表示4的右边有2个数比4小，则确定4的位置： _ 4 _ _
3. 第二个2表示3的右边有2个比3小，则确定3的位置：3 4 _ _
4. 第三个1表示2的右边有1个比2小，则确定2的位置：3 4 2 _
5. 最后确定1的位置：3 4 2 1

同时可以知道，3421这个排列的序号是2*3！+2*2！+1*1！+0=17。

而且每个序号对应唯一的排列，我们知道一个排列求序号就可以了，所以就愉快的解决了

然后浅谈一下我的意识流：

对于一个全排列数列，也就是1234-4321，按字典序依次。

还是对于3421，3后面有两个数比它小，1在第一位时有3！个全排列，2在第一位时有3！个全排列，所以3421在这个数列的位置一定在2*3！之后在3*3！之前。

4后面有2个比它小，因为3这一位固定了，那么它之前会有3 1 打头的2！个排列，3 2 打头的2！个全排列，它在2*3！+2*2！个数之后

2后面有1个比它小，3 4 2 都固定了，那么也就是3 4 2 打头的1！个全排列即它本身。

所以映射了唯一的编号，我们就可以做题了。

 

这道题输出格式两点注意：

1是字母60个一行要换行

2是它给出的串是顺时针念的（A sheet configuration is given by the sequence of colors obtained by reading the colors of the squares starting at the upper left corner and going in clockwise direction. For instance, the configuration of Figure 3 is given by the sequence (1,2,3,4,5,6,7,8). This configuration is the initial configuration.）

例如题目中的：

1234

8765

如果给出一个描述就是12345678

所以样例中的描述是

2684

1375

当然这样并不舒服，所以我们把它的格式强行改成两排横着念就好了

 /*
 ID: ivorysi
 PROG: msquare
 LANG: C++
 */
 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #include <queue>
 #include <set>
 #include <vector>
 #define siji(i,x,y) for(int i=(x);i<=(y);++i)
 #define gongzi(j,x,y) for(int j=(x);j>=(y);--j)
 #define xiaosiji(i,x,y) for(int i=(x);i<(y);++i)
 #define sigongzi(j,x,y) for(int j=(x);j>(y);--j)
 #define inf 0x7fffffff
 #define MAXN 400005
 #define ivorysi
 #define mo 97797977
 #define ha 974711
 #define ba 47
 #define fi first
 #define se second
 //#define pis pair<int,string>
 using namespace std;
 typedef long long ll;
 int fac[]={,,,,,,,,};
 int num(int x) {
     int ss[],l=;
     while(x>) {ss[++l]=x%;x/=;}
     int ret=;
     siji(i,,l) {
         int x=;
         xiaosiji(j,,i) {
             if(ss[j]<ss[i]) ++x;
         }
         ret+=x*fac[i];
     }
     return ret;
 }
 int cha1(int x) {
     int ss[],l=;
     while(x>) {ss[++l]=x%;x/=;}
     siji(i,,) swap(ss[i],ss[i+]);
     int ret=;
     gongzi(i,,) ret=ret*+ss[i];
     return ret;
 }
 int cha2(int x) {
     int ss[],l=;
     while(x>) {ss[++l]=x%;x/=;}
     int ss1[];
     siji(i,,) swap(ss[i],ss[-i+]);
     ss1[]=ss[];
     siji(i,,) ss1[i]=ss[i-];
     ss1[]=ss[];
     siji(i,,) ss1[i]=ss[i-];
     int ret=;
     siji(i,,) ret=ret*+ss1[i];
     return ret;
 }
 int cha3(int x) {
     int ss[],l=;
     while(x>) {ss[++l]=x%;x/=;}
     siji(i,,) swap(ss[i],ss[-i+]);
     int t=ss[];
     ss[]=ss[];ss[]=ss[];ss[]=ss[];ss[]=t;
     int ret=;
     siji(i,,) ret=ret*+ss[i];
     return ret;
 }
 int ys[];
 int mut[],step[],prev[],ans[],cnt;
 int to;
 char *str="$ABC";
 queue<int> q;
 void bfs() {
     q.push();
     mut[num()]=;
     while(!q.empty()) {
         int now=q.front();q.pop();
         if(now==to) break;
         int z=cha1(now);
         int w=num(z);
         if(mut[w]==) {
             mut[w]=mut[num(now)]+;
             step[w]=;
             prev[w]=num(now);
             q.push(z);
         }
         z=cha2(now);
         w=num(z);
         if(mut[w]==) {
             mut[w]=mut[num(now)]+;
             step[w]=;
             prev[w]=num(now);
             q.push(z);
         }
         z=cha3(now);
         w=num(z);
         if(mut[w]==) {
             mut[w]=mut[num(now)]+;
             step[w]=;
             prev[w]=num(now);
             q.push(z);
         }
     }

 }
 void solve() {

     siji(i,,) {
         scanf("%d",&ys[i]);
     }
     siji(i,,) to=to*+ys[i];
     gongzi(i,,) to=to*+ys[i];
     bfs();
     printf("%d\n",mut[num(to)]-);
     int k=num(to);
     while(step[k]!=) {
         ans[++cnt]=step[k];
         k=prev[k];
     }
     gongzi(i,cnt,) {
         printf("%c",str[ans[i]]);
         if((cnt-i+)%==) puts("");
     }
     puts("");
 }
 int main(int argc, char const *argv[])
 {
 #ifdef ivorysi
     freopen("msquare.in","r",stdin);
     freopen("msquare.out","w",stdout);
 #else
     freopen("f1.in","r",stdin);
 #endif
     solve();
 }

哦还想起来一个老师让我刷但是我特别不想刷的一个界面特别不友好特别有那种盗版的感觉但是老师说它好的oj上有这道题。

一段很长的空白之后……

出题人是没有刷过USACO吗……

虽然有点小开心呢但还是要说出题人别闹了这道题还是挺水的如果知道中介数的话……