Codeforces Round #324 (Div. 2) E. Anton and Ira 贪心

E. Anton and Ira

Time Limit: 1 Sec

Memory Limit: 256 MB

题目连接

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

Description

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 p_i and p_j, 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 p_i and p_j.

It is guaranteed that the solution exists.

Sample Input

4
4 2 1 3
3 2 4 1

Sample Output

HINT

题意

给你两个数组，你要求让第一个数组变成第二个数组，你可以进行交换操作

每次交换操作的代价是abs(i-j)

然后让你把方案输出出来

题解：

贪心题，你其实可以把每次交换想成线段一样的东西

你只要把这些线段压成一条线段就好了

------->
<-------------
----------------->
----------->
<---------------

直接暴力交换那些和自己有重叠的线段端点就好了

代码：

#include<iostream>
#include<math.h>
#include<string.h>
#include<vector>
#include<queue>
#include<stdio.h>
using namespace std;

int a[2005];
int b[2005];
int c[2005];
int d[2005];
int ans1[520005];
int ans2[520005];
int main()
{
    int n;
    int tot = 0;
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
        scanf("%d",&a[i]);
    for(int i=1;i<=n;i++)
        scanf("%d",&b[i]),c[b[i]]=i;
    int ans= 0;
    for(int i=1;i<=n;i++)
    {
        if(a[i]!=b[i]){
        int k = -1;
        for(int j=i+1;j<=n;j++)
            if(a[j]==b[i])
            {
                k = j;
                break;
            }
        while(k!=i)
        {
            for(int j = i;j<k;j++)
            {
                if(c[a[j]]>=k)
                {
                    ans += fabs(k-j);
                    swap(k,j);
                    swap(a[k],a[j]);
                    ans1[tot]=k;
                    ans2[tot]=j;
                    tot++;
                    break;
                }
            }
        }
        }
    }
    cout<<ans<<endl;
    cout<<tot<<endl;
    for(int i=0;i<tot;i++)
        printf("%d %d
",ans1[i],ans2[i]);
}