CF Round#324 (Div2) E. Anton and Ira

E. Anton and Ira
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
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.
解析
构造，贪心。
这个题做了好久！不开心！
思路还是比较显然，就是细节上容易写错，烦。
不想写题解了，就给点思路吧。

#include<cstdio>#include<stack>using namespace std;int N,K,P[2100],S[2100],ans[(int)4e6+100],cost=0;//hash[i]->num i is at pos hash[i] in Pint abs(int a){return a>0?a:-a;}void swap(int i,int j){    cost+=abs(i-j);    ans[++K]=i;ans[++K]=j;    P[i]^=P[j];P[j]^=P[i];P[i]^=P[j];}stack<int> box;int main(){    scanf("%d",&N);    for(int i=1;i<=N;i++) scanf("%d",P+i);    for(int i=1;i<=N;i++)     {int tmp; scanf("%d",&tmp); S[tmp]=i;}    for(int i=1;i<=N;i++) P[i]=S[P[i]];    for(int i=1;i<=N;i++)    {        if(P[i]>i) box.push(i);        int pos=i;        while(box.size() && P[pos]<=box.top() && P[box.top()]>=pos)        {            swap(box.top(),pos);            i=pos-1;//等会循环结束了还有i++            pos=box.top();box.pop();            /*            为什么要运动i的位置？            1.            因为交换了box.top()和pos之后，            不能确定box.top()是不是已经到该到的位置了（或者说写起来很烦）            所以就干脆把i直接弄到pos（i=pos-1;//等会循环结束了还有i++）            2.            可以将改动后数列再次扫描，避免遗漏。（非常重要，不然算法有漏洞）            */        }        //for(int j=1;j<=N;j++) printf("%d%c",P[j],j==N?10:32);    }    printf("%d\n%d\n",cost,K>>1);    for(int i=1;i<=K;i++) printf("%d%c",ans[i],i&1?32:10);}