NJUST 1925 sequence

原题

Description

将一个给定的数列，拆分成K个不降序列，每个数出现且只出现一次，且在各序列中各个数相对于原数列的相对顺序不变。

如7 6 9 8 10可以拆成 7 9 10和6 8。求最小的K值。

Algorithm

按顺序生产

比如先 7 9 10

再 6 8

然后拿链表实现

Code

#include <cstdio>
#include <iostream>
#include <list>
using namespace std;
void solve()
{
list<int> mylist;
int n;
scanf("%d", &n);
for (int i = 0; i < n; i++)
{
int x;
scanf("%d", &x);
mylist.push_back(x);
}
int ans = 0;
while (!mylist.empty())
{
int head = *mylist.begin();
mylist.pop_front();
//   for (list<int>::iterator it = mylist.begin(); it != mylist.end(); it++)
//   cout << *it;
for (list<int>::iterator it = mylist.begin(); it != mylist.end(); it++)
if (*it >= head)
{
head = *it;
it = mylist.erase(it);
it--;
}
ans++;
}
cout << ans << endl;
}
int main()
{
//  freopen("input.txt", "r", stdin);
int t = 1;
scanf("%d", &t);
for (int i = 0; i < t; i++)
solve();
}