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;
}