POJ 3670

最新推荐文章于 2019-07-23 13:23:43 发布

原创最新推荐文章于 2019-07-23 13:23:43 发布 · 222 阅读

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本文介绍了一个有趣的算法问题：如何通过最小数量的操作使牛群按晚餐组别升序或降序排列。采用最长非严格递增子序列（LIS）算法解决此问题，并提供了一段简洁高效的C++代码实现。

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Description
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.

Input
* 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

Output
* 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

Sample Input
5
1
3
2
1
1
Sample Output
1

题目：
求一个最长非严格子串的长度，剩下的就是要改的。就是求LIS。
这是学长的代码，用到了stl的东西。

#include <iostream>
#include <cstring>
#include <string>
#include <algorithm>
#include <cstdio>
#include <map>
using namespace std;
const int INF = 0x3f3f3f3f;
int a[300010];int b[300010];
int n;
int judge()
{
    for(int i = 0; i < n; i++) b[i] = INF;
    for(int i = 0; i < n; i++)
        *upper_bound(b, b + n, a[i]) = a[i];
    return lower_bound(b, b+n, INF) - b;
}
int main()
{
    scanf("%d", &n);
    for(int i = 0; i < n; i++)
        scanf("%d", &a[i]);
    int ans = INF;
    ans = min(ans, n - judge());
    reverse(a, a + n);
    ans = min(ans, n - judge());
    printf("%d\n", ans);
    return 0;
}