USACO 2008 FEB Eating Together

题目描述

The cows are so very silly about their dinner partners. They have organized themselves into three groups (conveniently numbered 1, 2, and 3) that insist upon dining together. The trouble starts when they line up at the barn to enter the feeding area.

Each cow i carries with her a small card upon which is engraved Di (1 ≤ Di ≤ 3) indicating her dining group membership. The entire set of N (1 ≤ N ≤ 30,000) cows has lined up for dinner but it's easy for anyone to see that they are not grouped by their dinner-partner cards.

FJ's job is not so difficult. He just walks down the line of cows changing their dinner partner assignment by marking out the old number and writing in a new one. By doing so, he creates groups of cows like 111222333 or 333222111 where the cows' dining groups are sorted in either ascending or descending order by their dinner cards.

FJ is just as lazy as the next fellow. He's curious: what is the absolute mminimum number of cards he must change to create a proper grouping of dining partners? He must only change card numbers and must not rearrange the cows standing in line.

FJ的奶牛们在吃晚饭时很傻。他们把自己组织成三组（方便编号为1, 2和3），坚持一起用餐。当他们在谷仓排队进入喂食区时，麻烦就开始了。

每头奶牛都随身带着一张小卡片，小卡片上刻的是Di（1≤Di≤3）表示她属于哪一组。所有的N（1≤N≤30000）头奶牛排队吃饭，但他们并不能按卡片上的分组站好。

FJ的工作并不是那么难。他只是沿着牛的路线走下去，把旧的号码标出来，换上一个新的。通过这样做，他创造了一群奶牛，比如111222333或333222111，奶牛的就餐组按他们的晚餐卡片按升序或降序排列。

FJ就像任何人一样懒惰。他很好奇：怎样他才能进行适当的分组，使得他只要修改最少次数的数字？由于奶牛们已经很长时间没有吃到饭了，所以“哞哞”的声音到处都是，FJ只好更换卡号，而不能重新排列已经排好队的奶牛。

输入输出格式

输入格式：

* Line 1: A single integer: N

* Lines 2..N+1: Line i describes the i-th cow's current dining group with a single integer: Di

• 第1行：一个整数：n

• 第2~n+1行：第i-1行描述第i个奶牛目前分组。

输出格式：

* Line 1: A single integer representing the minimum number of changes that must be made so that the final sequence of cows is sorted in either ascending or descending order

一个整数，表示必须做出的最小变化数，以便以升序或降序排序最终序列。

输入输出样例

输入样例#1： 复制

5
1
3
2
1
1

输出样例#1： 复制

1 

说明

感谢@一思千年 提供翻译

分析

我们要找最少的更改吊牌的次数，其实也能放过来思考最多的不用更改吊牌的数量。由于奶牛们必须按照123或者321的顺序来排，那么如果存在一个不下降子序列或者不上升子序列，那么这个序列就是不用被调整的，而其余的需要进行调整。那么问题变成了求这个数列当中最长不下降子序列和最长不上升子序列的更大值。

接下来问题来了，如果按照我们平常写动规的方式来写这道题，就是标准的动规时间复杂度O(n^2)。这显然对我们不是很友好。如果不想写二分的最长不下降子序列，我们就要来观察这道题里的一些特殊点，如何从特殊点突破，来降低时间复杂度。那么就很显然了，这道题可能出现的数字只有：123。那么我们要做的就是像接下来程序里的那样：对于每一个数字，f[i][n]保存到当前位置，且当前最优子序列的最大数为n的情况下，序列的长度。

程序

#include <bits/stdc++.h>
using namespace std;
int n, a[30000 + 1], f[30000+1][4], ans = 0;
int main()
{
    cin >> n;
    for (int i = 1; i <= n; i++)
        cin >> a[i];
    for (int i = 1; i <= n; i++)
    {
        f[i][1] = f[i-1][1] + (a[i]==1);
        ans = max(ans,f[i][1]);
        f[i][2] = max(f[i-1][1],f[i-1][2])+(a[i]==2);
        ans = max(ans,f[i][2]);
        f[i][3] = max(f[i-1][1],max(f[i-1][2],f[i-1][3]))+(a[i]==3);
        ans = max(ans,f[i][3]);
    }
    memset(f,0,sizeof(f));
    for (int i = 1; i <= n; i++)
    {
        f[i][3] = f[i-1][3] + (a[i]==3);
        ans = max(ans,f[i][3]);
        f[i][2] = max(f[i-1][3],f[i-1][2])+(a[i]==2);
        ans = max(ans,f[i][2]);
        f[i][1] = max(f[i-1][1],max(f[i-1][2],f[i-1][3]))+(a[i]==1);
        ans = max(ans,f[i][1]);
    }
    cout << n-ans << endl;
    return 0;
}