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,,sizeof(mi));
     memset(mx,,sizeof(mx));
     memset(f,,sizeof(f));
     memset(r,,sizeof(r));
     for(int i=;i<=n;i++)
     {
         mi[i][i]=a[i];
         mx[i][i]=a[i];
         f[i][i]=;
         r[i]=i;
     }
     //为mi,mx赋值
     for(int i=;i<=n;i++)
     for(int j=i+;j<=n;j++)
     {
         mx[i][j]=max(a[j],mx[i][j-]);
         mi[i][j]=min(a[j],mi[i][j-]);
     }
     //为f数组赋值
     for(int t=;t<=n;t++)//t在枚举区间长度
     for(int i=;i+t-<=n;i++)
     {
         int j=i+t-;
         //不是连续的一段无法分区间
         if(mx[i][j]-mi[i][j]!=t-)
             f[i][j]=;
         else
         {
             //j一定大于r[i]
             if(mi[i][r[i]]>mi[i][j])
                 f[i][j]=;
             else
                 f[i][j]=f[i][r[i]]+f[r[i]+][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(,f[][n]);//WA点，一开始写成res=1就WA了
     //先枚举seg_a
     for(int i=;i<=n;i++)
     for(int j=i;j<=n;j++)
     {
         //满足条件才能继续枚举seg_b
         if(i==||(f[][i-]!=&&mi[][i-]==))
         {
             k=mx[i][j];
             if(f[i][j]&&(k==n||(f[k+][n]!=&&mx[k+][n]==n)))
             {
                 for(int t=j+;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[][i-]++f[j+][t-]++f[k+][n]);
                     }
                 }
             }
         }
     }
 }

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

 }