UVA6588 - Crane

​​Regionals 2013​​​ >> ​​​ Europe - Central​​

​​6588 - Crane​​

题意：给定一组无序数，从1-n，要求排序，排序的要求是选定一个区间[a，b]，则这个区间的前一半和后一半互换位置，a到b有偶数个数，问交换几次。

解题思路：由于数字个数不多，可以枚举1-n个位置，枚举第i个位置，就找到第i个点所在点的位置，如果可以直接交换就交换，如果不能就一半一半交换。

#include <stdio.h>
#include <iostream>
#include <string.h>
#include <math.h>
#include <algorithm>
#include <stdlib.h>
#include <queue>
#include <vector>
#include <map>

using namespace std;

const int inf = 0x3f3f3f3f;
const int Max = 10100;
typedef long long int LL;
int a[Max],vis[Max], n, rr[531500], ll[531500];
int cut;
void exch(int l, int r)
{
    int i, t;
    int mid = r-l;
    ll[cut] = l;
    rr[cut++] = r+mid-1;
    //printf("%d %d\n", l, r+mid-1);
    for(i = l; i < r; i++)
    {
        t = a[i];
        a[i] = a[i+mid];
        a[i+mid] = t;
        t = vis[a[i]];
        vis[a[i]] = vis[a[i+mid]];
        vis[a[i+mid]] = t;
    }
    /*for(i = 1; i <= n; i++)
    {
        cout<<a[i]<<" ";
    }
    cout<<endl;*/
}

int main()
{
    int i, pos, j;
    int K;
    scanf("%d", &K);
    while(K--)
    {
        scanf("%d",&n);
        cut = 0;
        for(i = 1; i <= n; i++)
        {
            scanf("%d",&a[i]);
            vis[a[i]] = i;
        }
        for(i = 1; i <= n; i++)
        {
            if(vis[i] == i)
                continue;
            pos = vis[i];
            while(a[i] != i)
            {
                int end = i+(pos-i)*2-1;
                if(end <= n)
                {
                    exch(i, pos);
                }
                else
                {
                    int mid = pos - i +1;
                    if(mid%2 == 0)
                    {
                        exch(i, i + mid/2);
                        pos = vis[i];
                    }
                    else
                    {
                        exch(i+1, i+mid/2+1);
                        pos = vis[i];
                    }
                }
            }
        }
        printf("%d\n", cut);
        for(i = 0; i < cut; i++)
        {
            printf("%d %d\n", ll[i], rr[i]);
        }
    }
    return 0;
}