codeforces 324# E. Anton and Ira (暴力枚举+贪心)

最新推荐文章于 2019-10-28 00:40:19 发布

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最新推荐文章于 2019-10-28 00:40:19 发布

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分类专栏： ACM-贪心 ACM-暴力枚举文章标签： codefroces 枚举贪心

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本文解析了Codeforces上的一道经典题目，通过寻找最优交换策略将一个排列转换为另一个排列，以最小化交换成本。文章提供了详细的算法思路及C++实现代码。

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题目：http://codeforces.com/contest/584/problem/E

题意：给定两个排列p1和p2。可以交换p1中的两个元素p1[i],p1[j]，花费为|i-j|，求最小的话费使得p1变为p2。

Anton loves transforming one permutation into another one by swapping elements for money, and Ira doesn't like paying for stupid games. Help them obtain the required permutation by paying as little money as possible.

More formally, we have two permutations, p and s of numbers from 1 to n. We can swap pi and pj, by paying |i - j| coins for it. Find and print the smallest number of coins required to obtain permutation s from permutation p. Also print the sequence of swap operations at which we obtain a solution.

Input

The first line contains a single number n (1 ≤ n ≤ 2000) — the length of the permutations.

The second line contains a sequence of n numbers from 1 to n — permutation p. Each number from 1 to n occurs exactly once in this line.

The third line contains a sequence of n numbers from 1 to n — permutation s. Each number from 1 to n occurs once in this line.

Output

In the first line print the minimum number of coins that you need to spend to transform permutation p into permutation s.

In the second line print number k (0 ≤ k ≤ 2·106) — the number of operations needed to get the solution.

In the next k lines print the operations. Each line must contain two numbers i and j (1 ≤ i, j ≤ n, i ≠ j), which means that you need to swap pi and pj.

It is guaranteed that the solution exists.

        Sample test(s) 
      

          input 
        
4
4 2 1 3
3 2 4 1

          output 
        
3
2
4 3
3 1

Note

In the first sample test we swap numbers on positions 3 and 4 and permutation p becomes 4 2 3 1. We pay |3 - 4| = 1 coins for that. On second turn we swap numbers on positions 1 and 3 and get permutation 3241 equal to s. We pay |3 - 1| = 2 coins for that. In total we pay three coins.

分析：记p1[i]在p2中的位置为pos1，p1[j]在p2中的位置为pos2，要把p1[i]放到pos1位置，很明显，当pos1<=j而且po2<=i时p1[i]与p2[j]交换的性价比最高。然后3层循环枚举，一直交换，直到某个时刻数组扫了一遍没有元素交换，那么p1就变成了p2。

ps:居然过了。。。

代码：

#include <iostream>
#include <cstdio>
#include <map>
using namespace std;
const int maxn = 3000;
int a[maxn],b[maxn];

int mp[3000000];
struct node
{
	int x,y;
	node()=default;
	node(int a,int b)
	:x(a),y(b){}
}s[3000000];
int Abs(int x)
{
	if(x<0)
		return -x;
	return x;
}
int main()
{
	int n,i,j,p=0,ans=0;
	cin>>n;
	for(i=1;i<=n;i++)
		cin>>a[i];
	for(i=1;i<=n;i++)
	{
		cin>>b[i]; 
		mp[b[i]]=i;
	}
	int cur,temp;
	while(true)
	{
		int fg2=0;
		for(i=1;i<=n;i++)
		{
			temp=mp[a[i]];
			if(temp==i)
				continue ;
			int cur=i,goal=temp;
			while(cur<goal)
			{
				int fg1=0;
				for(j=cur+1;j<=n && j<=goal;j++)
				{
					temp=mp[a[j]];
					if(temp<=cur)
					{
						ans+=Abs(cur-j);
						s[p++]=node(cur,j);
						swap(a[cur],a[j]);
						cur=j;
						fg2=fg1=1;
						break;
					}
				}
				if(!fg1)
					break;
			}
		}
		if(!fg2)
			break;
	}
	printf("%d\n",ans);
	printf("%d\n",p);
	for(i=0;i<p;i++)
		printf("%d %d\n",s[i].x,s[i].y);
	return 0;
}