A permutation p of length n is a sequence of distinct integers p₁, p₂, ..., p_n (1 ≤ p_i ≤ n). A permutation is an identity permutation, if for any i the following equation holds p_i = i.

A swap (i, j) is the operation that swaps elements p

Valera wonders, how he can transform permutation p into any permutation q, such that f(q) = m, using the minimum number of swaps. Help him do that.

The first line contains integer n (1 ≤ n ≤ 3000) — the length of permutation p. The second line contains n distinct integers p₁, p₂, ..., p_n (1 ≤ p_i ≤ n) — Valera‘s initial permutation. The last line contains integer m

In the first line, print integer k — the minimum number of swaps.

In the second line, print 2k integers x₁, x₂, ..., x_2k — the description of the swap sequence. The printed numbers show that you need to consecutively make swaps (x₁, x₂), (x₃, x₄), ..., (x_2k - 1, x_2k).

If there are multiple sequence swaps of the minimum length, print the lexicographically minimum one.

5
1 2 3 4 5
2

2
1 2 1 3 

5
2 1 4 5 3
2

1
1 2 

Sequence x₁, x₂, ..., x_s is lexicographically smaller than sequence y₁, y₂, ..., y_s, if there is such integer r (1 ≤ r ≤ s), that x₁ = y₁, x₂ = y₂, ..., x_r - 1 = y_r - 1 and x_r < y_r.

題意：有函數f(p)，表示把排列p，變為順序的排列，所用的最小交換次數，這裏的交換是兩兩交換，而不是相鄰交換。

現在給你排列p和m，讓你交換最小的次數，使得排列變為q，滿足f(q)=m，如果有多種交換方案，輸出字典序最小的;

思路：1，把排列p變為1,2,3....p，的最小次數=N-環數。假如N-環數<m，那就加環。 否則減環。

2，如果交換兩個不在同一個環的位置的數，那麽環數+1；

3，如果交換在同一個環的位置的數，那麽環數-1；

所以我們的任務就是加環或者減環，而每次分組的復雜度是O(N)；我們最多加N次環，所以我們可以暴力交換，交換後重新分組。

復雜度O(N^2)；

#include<bits/stdc++.h>
#define rep(i,a,b) for(int i=a;i<=b;i++)
using namespace std;
const int maxn=3010;
int a[maxn],vis[maxn],tot,N,M;
void dfs(int u)
{
    if(vis[u]) return ;
    vis[u]=tot;
    dfs(a[u]);
}
void rebuild()
{
    tot=0; rep(i,1,N) vis[i]=0;
    rep(i,1,N) if(!vis[i]) tot++,dfs(i);
}
int main()
{
    scanf("%d",&N);
    rep(i,1,N) scanf("%d",&a[i]);
    rebuild();
    scanf("%d",&M);
    if(M==N-tot) return puts("0"),0;
    int ans,opt=0;
    if(M>N-tot) ans=M-N+tot,opt=1; //合
    else ans=N-tot-M;//拆
    printf("%d\n",ans);
    rep(i,1,N)
     rep(j,1,N){
        if(!ans) break;
        if(i!=j&&(abs(vis[i]-vis[j])?1:0)==opt) {
            swap(a[i],a[j]); ans--;
            printf("%d %d ",i,j);
            rebuild();
        }
    }
    return 0;
}

CodeForces - 441D： Valera and Swaps（置換群）