HDU 6049 - Sdjpx Is Happy | 2017 Multi-University Training Contest 2

/*
HDU 6049 - Sdjpx Is Happy [ 枚举,剪枝 ]  |  2017 Multi-University Training Contest 2
题意：
	长度为N的排列 N <= 3000
	排序分三个步骤：
		1.原数组分为不相交的K段
		2.每段都独立排序
		3.选择其中两段swap
	问按步骤能成功排序的K能取到的最大是多少
分析：
	先预处理出任意段的最小值和最大值
	再处理出任意[l,r]段最多能分成多少段有效段，用f[i,j]表示
	所谓的有效段首先满足 Max[i,j]-Min[i,j] = j-i
	再满足其中每段依此递增，即前一段的最大值 == 后一段的最小值-1
	设需要交换的前一段为[i, j] 后一段为[k, t]
	枚举i,j，则 t = Max[i][j]，再枚举k，更新答案
	虽然枚举复杂度大，但可以剪枝
	比如要求：f[i,j] > 0 && i 如果不为 1 则 Min[1,i-1] = 1, Max[1, i-1] = i-1
	类似的对k, t剪枝
*/
#include <bits/stdc++.h>
using namespace std;
const int N = 3005;
int Max[N][N], Min[N][N];
int f[N][N];
int t, n;
int a[N], last[N];
void init()
{
    memset(f, 0, sizeof(f));
    int i, j, k;
    for (i = 1; i <= n; i++) Max[i][i] = Min[i][i] = a[i];
    for (k = 2; k <= n; k++)
        for (i = 1; i+k-1 <= n; i++)
        {
            j = i+k-1;
            Max[i][j] = max(Max[i+1][j], Max[i][i]);
            Min[i][j] = min(Min[i+1][j], Min[i][i]);
        }
    for (i = 1; i <= n; i++) f[i][i] = 1, last[i] = i;
    for (k = 2; k <= n; k++)
        for (i = 1; i+k-1 <= n; i++)
        {
            j = i+k-1;
            if (Max[i][j] - Min[i][j] != j-i) continue;
            if (Min[i][last[i]] != Min[i][j]) f[i][j] = 1;
            else f[i][j] = f[i][last[i]] + 1;
            last[i] = j;
        }
}
int ans;
void solve()
{
    ans = f[1][n];
    for (int i = 1; i <= n; i++)
    {
        if ( i != 1 && (!f[1][i-1] || Max[1][i-1] != i-1)) continue;
        for (int j = i; j <= n; j++)
        {
            if (!f[i][j]) continue;
            int t = Max[i][j];
            if (t != n && (!f[t+1][n] || Min[t+1][n] != t+1 || Max[t+1][n] != n)) continue;
            for (int k = t; k > j; k--)
            {
                if (!f[k][t] || Min[k][t] != i ) continue;
                if (k > j+1)
                {
                   if (!f[j+1][k-1] || Max[k][t] != Min[j+1][k-1]-1 || Min[i][j] != Max[j+1][k-1]+1) continue;
                }
                else
                {
                    if (Max[k][t] != Min[i][j]-1) continue;
                }
                ans = max(ans, f[1][i-1] + 2 + f[j+1][k-1] + f[t+1][n]);
            }
        }
    }
}
int main()
{
    scanf("%d", &t);
    while (t--)
    {
        scanf("%d", &n);
        for (int i = 1; i <= n; i++) scanf("%d", a+i);
        init();
        solve();
        printf("%d\n", ans);
    }
}