Heap 3214 LIS题解

依据问题转换成最长不降子序列问题。

10^9的输入数据计算起来还是挺花时间的。由于这里仅仅能使用O(nlgn)时间复杂度了。

只是证明是能够算出10^9个数据的。

由于时间限制是5s.

#include <stdio.h>
#include <vector>
#include <string.h>
#include <algorithm>
#include <iostream>
#include <string>
#include <limits.h>
#include <stack>
#include <queue>
#include <set>
#include <map>
using namespace std;

const int MAX_N = 20;
vector<int> arr, a2;
int N;

inline int lSon(int rt) { return rt<<1|1; }
inline int rSon(int rt) { return (rt<<1)+2; }

void postOrder(int rt, int &v)
{
	int l = lSon(rt), r = rSon(rt);
	if (l < N) postOrder(l, v);
	if (r < N) postOrder(r, ++v);
	a2.push_back(arr[rt]-v);
}

int biGetIndex(int low, int up, int v)
{
	while (low <= up)
	{
		int mid = low + ((up-low)>>1);
		if (v < a2[mid]) up = mid-1;
		else low = mid+1;
	}
	return low;
}

int LIS()
{
	int j = 0;
	for (int i = 1; i < N; i++)
	{
		if (a2[i] >= a2[j]) a2[++j] = a2[i];
		else
		{
			int id = biGetIndex(0, j, a2[i]);
			a2[id] = a2[i];
		}
	}
	return j+1;
}

int main()
{
	int a;
	scanf("%d", &N);
	arr.clear(), a2.clear();
	while (scanf("%d", &a) != EOF)
	{
		arr.push_back(a);
	}

	N = (int) arr.size();
	int v = 0;
	postOrder(0, v);
	int len = LIS();
	printf("%d\n", N-len);
	return 0;
}