HDU 6049 17多校2 Sdjpx Is Happy（思维题difficult）

Problem Description

Sdjpx is a powful man,he controls a big country.There are n soldiers numbered 1~n(1<=n<=3000).But there is a big problem for him.He wants soldiers sorted in increasing order.He find a way to sort,but there three rules to obey.
1.He can divides soldiers into K disjoint non-empty subarrays.
2.He can sort a subarray many times untill a subarray is sorted in increasing order.
3.He can choose just two subarrays and change thier positions between themselves.
Consider A = [1 5 4 3 2] and P = 2. A possible soldiers into K = 4 disjoint subarrays is:A1 = [1],A2 = [5],A3 = [4],A4 = [3 2],After Sorting Each Subarray:A1 = [1],A2 = [5],A3 = [4],A4 = [2 3],After swapping A4 and A2:A1 = [1],A2 = [2 3],A3 = [4],A4 = [5].
But he wants to know for a fixed permutation ,what is the the maximum number of K?
Notice: every soldier has a distinct number from 1~n.There are no more than 10 cases in the input.

 

 

Input

First line is the number of cases.
For every case:
Next line is n.
Next line is the number for the n soildiers.

 

 

Output

the maximum number of K.
Every case a line.

 

 

Sample Input

2

5

1 5 4 3 2

5

4 5 1 2 3

 

 

Sample Output

4

2

Hint

Test1: Same as walk through in the statement. Test2: [4 5] [1 2 3] Swap the 2 blocks: [1 2 3] [4 5].

 

启发博客：http://www.cnblogs.com/FxxL/p/7253028.html

题意： 给出n，一个1~n的排列，要求进行三步操作

           1.分区（随便分）

           2.对分好的每一个区内进行从小到大的排序

           3.挑选两个区进行交换（只能挑选两个，只能交换易次），使得序列的顺序变成1-n;

          问在满足要求的情况下，最多能分成多少区

题解：第一步是分区，第二步是枚举。

          分区是开了一个f[i][j]数组用来记录，i-j区间里可以有多少满足要求的段，用到mx,mi，r来辅助，具体可见代码注释。

          枚举是枚举要交换的两个区间，每次更新答案的最大值。设左右区间分别为seg_a,seg_b。

          seg_a要满足：第一段或者之前包括1~i-1的所有数字，当然自身不能为空

                                把这个区间最大的数定义为k，根据k来枚举seg_b，k是seg_b的右端点

                                还需满足k==n或者k+1及以后的所有数字包含k+1~n

          seg_b要满足：k是seg_b的右端点，自身不为空，要保证它的最小值是i

          像上述这样来做即可，当然中间会有一些不小心造成的WA点，大家注意即可

 

#include<cstdio>
#include<iostream>
#include<cmath>
#include<cstring>
using namespace std;
#define MAXN 3005

int a[MAXN],res,n;
int mi[MAXN][MAXN],mx[MAXN][MAXN];
//mi[i][j]表示从i到j的最小值，mx[i][j]表示从i到j的最大值
int f[MAXN][MAXN],r[MAXN];
//f[i][j]表示从i到j可以分成的区间数，r[i]表示最近一次从i开始的区间的右端（方便更新）


void init()//第一步，分块
{
    memset(mi,0,sizeof(mi));
    memset(mx,0,sizeof(mx));
    memset(f,0,sizeof(f));
    memset(r,0,sizeof(r));
    for(int i=1;i<=n;i++)
    {
        mi[i][i]=a[i];
        mx[i][i]=a[i];
        f[i][i]=1;
        r[i]=i;
    }
    //为mi,mx赋值
    for(int i=1;i<=n;i++)
    for(int j=i+1;j<=n;j++)
    {
        mx[i][j]=max(a[j],mx[i][j-1]);
        mi[i][j]=min(a[j],mi[i][j-1]);
    }
    //为f数组赋值
    for(int t=2;t<=n;t++)//t在枚举区间长度
    for(int i=1;i+t-1<=n;i++)
    {
        int j=i+t-1;
        //不是连续的一段无法分区间
        if(mx[i][j]-mi[i][j]!=t-1)
            f[i][j]=0;
        else
        {
            //j一定大于r[i]
            if(mi[i][r[i]]>mi[i][j])
                f[i][j]=1;
            else
                f[i][j]=f[i][r[i]]+f[r[i]+1][j];
            r[i]=j;//这个r数组很精华
        }
    }
}

void solve()//第二步，枚举找交换区间
{
    int k;
    res=max(1,f[1][n]);//WA点，一开始写成res=1就WA了
    //先枚举seg_a
    for(int i=1;i<=n;i++)
    for(int j=i;j<=n;j++)
    {
        //满足条件才能继续枚举seg_b
        if(i==1||(f[1][i-1]!=0&&mi[1][i-1]==1))
        {
            k=mx[i][j];
            if(f[i][j]&&(k==n||(f[k+1][n]!=0&&mx[k+1][n]==n)))
            {
                for(int t=j+1;t<=k;t++)
                {
                    if(f[t][k]&&mi[t][k]==i)
                    {
                        //printf("%d %d %d %d %d\n",i,j,t,k,f[1][i-1]+1+f[j+1][t-1]+1+f[k+1][n]);
                        res=max(res,f[1][i-1]+1+f[j+1][t-1]+1+f[k+1][n]);
                    }
                }
            }
        }
    }
}

int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
            scanf("%d",&a[i]);
        init();
        solve();
        printf("%d\n",res);
    }
    return 0;

}