USACO-Section 3.2-PROB Magic Squares

深入探讨由Mr. Rubik发明的平面版魔方——魔方格(MagicSquares),以及如何通过三种基本变换将其从初始配置转换为目标配置。本文详细介绍了变换规则、输入输出格式,并提供了实现该任务的C++代码片段,旨在求解最小序列的基本变换。此教程特别适合热衷于算法挑战和编程技巧的读者。
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:

1234
8765

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:
8765
1234
B:
4123
5876
C:
1724
8635

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

搜索题。注意对string和map的运用
要给string的某一位赋值,前提是它已经被赋值

#include<cstdio>
#include<cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<map>
#include<queue>
#define name "msquare"
using namespace std;
map <string,int> num;
struct father
{
	int f;//是由谁变来的 
	char c;//变的方式 
}fa[1000000];
queue <string> Q;
string s,t;
char ABC[100000];
int ans,tot;
void print()
{
	int x=0,y=tot+1;
	while (y!=0)
	{
		ABC[++x]=fa[y].c;
		y=fa[y].f;
	}
	cout<<x<<endl;
	int tmp;
	for (int i=x;i>=1;i--)
	{
		tmp++;
	    cout<<ABC[i];
	    if (tmp==60) cout<<endl;
    }
	cout<<endl;
}
void bfs()
{
	int i,j;
	while (!Q.empty())
	{
		string now=Q.front();
		Q.pop();
		string tmp="00000000";
		for (i=0;i<=7;i++)
		tmp[i]=now[7-i];
		//cout<<tmp<<endl;
		if (string(t)==string(tmp)) {fa[tot+1].f=num[now];fa[tot+1].c='A';print();break;}
		if (num[tmp]==0) 
		{
			Q.push(tmp);
			++tot;
			num[tmp]=tot;
			fa[tot].f=num[now];
			fa[tot].c='A';
		}
		tmp[0]=now[3];tmp[7]=now[4];
		tmp[1]=now[0];tmp[2]=now[1];tmp[3]=now[2];
		tmp[6]=now[7];tmp[5]=now[6];tmp[4]=now[5];
		//cout<<tmp<<endl;
		if (string(t)==string(tmp)) {fa[tot+1].f=num[now];fa[tot+1].c='B';print();break;}
		if (num[tmp]==0) 
		{
			Q.push(tmp);
			++tot;
			num[tmp]=tot;
			fa[tot].f=num[now];
			fa[tot].c='B';
		}
		tmp=now;
		tmp[1]=now[6];tmp[2]=now[1];
		tmp[5]=now[2];tmp[6]=now[5];
		//cout<<tmp<<endl;
		if (string(t)==string(tmp)) {fa[tot+1].f=num[now];fa[tot+1].c='C';print();break;}
		if (num[tmp]==0) 
		{
			Q.push(tmp);
			++tot;
			num[tmp]=tot;
			fa[tot].f=num[now];
			fa[tot].c='C';
		}
	}
}
int main()
{
	freopen(name ".in","r",stdin);
	freopen(name ".out","w",stdout);
	int i,j;
	t="00000000";
	for (i=0;i<=7;i++)
	    cin>>t[i];
	//cout<<t;
    s="12345678";
    if (t=="12345678") {cout<<"0";cout<<endl<<endl;return 0;}
    num[s]=0;
    Q.push(s);
    bfs();
	return 0;
}


### USACO 1327 Problem Explanation USACO 1327涉及的是一个贪心算法中的区间覆盖问题。具体来说,这个问题描述了一组奶牛可以工作的班次范围,并要求找出最少数量的奶牛来完全覆盖所有的班次。 对于此类问题的一个有效方法是采用贪心策略[^1]。首先按照区间的结束时间从小到大排序这些工作时间段;如果结束时间相同,则按开始时间从早到晚排列。接着遍历这个有序列表,在每一步都尽可能选择最早能完成当前未被覆盖部分的工作时段。通过这种方式逐步构建最终解集直到所有的时间段都被覆盖为止。 为了提高效率并防止超时错误,建议使用`scanf()`函数代替标准输入流操作符`cin`来进行数据读取处理[^2]。 ```cpp #include <iostream> #include <vector> #include <algorithm> using namespace std; struct Interval { int start; int end; }; bool compareIntervals(const Interval& i1, const Interval& i2) { return (i1.end < i2.end || (i1.end == i2.end && i1.start < i2.start)); } int main() { vector<Interval> intervals = {{1, 7}, {3, 6}, {6, 10}}; sort(intervals.begin(), intervals.end(), compareIntervals); int currentEnd = 0; int count = 0; for (const auto& interval : intervals) { if (interval.start > currentEnd) break; while (!intervals.empty() && intervals.front().start <= currentEnd) { if (intervals.front().end >= interval.end) { interval = intervals.front(); } intervals.erase(intervals.begin()); } currentEnd = interval.end; ++count; if (currentEnd >= 10) break; // Assuming total shift length is known. } cout << "Minimum number of cows needed: " << count << endl; } ``` 此代码片段展示了如何实现上述提到的方法解决该类问题。需要注意的是实际比赛中可能还需要考虑更多边界条件以及优化细节以满足严格的性能需求。
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