CodeForces_1437E Make It Increasing(动态规划)_you are given an array of n integers 1 , 2 ,

Make It Increasing

time limit per test：2 seconds memory limit per test：256 megabytes

Problem Description

You are given an array of n integers $a_1,a_2,...,a_n$, and $a$ set $b$ of $k$ distinct integers from $1$ to $n$.

In one operation, you may choose two integers $i$ and $x$ ($1\le i\le n$, $x$ can be any integer) and assign $a_i:=x$. This operation can be done only if $i$ does not belong to the set $b$.

Calculate the minimum number of operations you should perform so the array a is increasing (that is, $a_1<a_2<a_3<\cdots <a_n$), or report that it is impossible.

Input

The first line contains two integers $n$ and $k$ ($1\le n\le 5\cdot 10^5,0\le k\le n$) — the size of the array a and the set b, respectively.

The second line contains $n$ integers $a_1,a_2,...,a_n$ ($1\le a_i\le 10^9$).

Then, if $k\ne 0$, the third line follows, containing $k$ integers $b_1,b_2,...,b_k$ ($1\le b_1<b_2<\cdots <b_k\le n$). If $k=0$, this line is skipped.

Output

If it is impossible to make the array a increasing using the given operations, print $-1$.

Otherwise, print one integer — the minimum number of operations you have to perform.

Sample Input

7 2
1 2 1 1 3 5 1
3 5

Sample Output

4

题意

有一个数组a，数组某些位置的数不可操作，其余位置的数可以进行操作，将其赋为任意值。求最少的操作次数，使数组a为升序序列，或者输出无解。

思路

判断无解：因为要求序列a严格递增，所以若 两个相邻的不可动的位置的值的差 小于 他们之间的距离，则无解。即$a_{pos2}-a_{pos1}<pos2-pos1$时无解($pos1<pos2$)。

最少操作次数：有些位置的数不能更改，则这些位置将数组分割为若干段，取其中连续的一段为，$d_1,d_2,...d_k$，其中$d_1,d_k$是固定的(为方便计算，可以另$a_0=-INF,a_{n+1}=INF$)。
由样例可知单纯的求最长上升子序列是不行的。考虑转化一下，设$e_1=0,e_i=d_i-d_1-(i-1)$，e代表若当前位置的数不变，$e_i$与$e_1$的最小差值。(因为首位不可变，所以当$e_i<0$时，可以考虑将其赋为INF)。
现在问题就转化为求以$e_k$结尾的，最长不降子序列的长度len。则 区间长度-len 为另当前子段单调递增所需要的最少操作次数。
所有子段的结果相加，即为所求。

#include<cstdio>
#include<iostream>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<map>
#include<vector>
#include<queue>
#include<iterator>
#include<string>
#define dbg(x) cout<<#x<<" = "<<x<<endl;
#define INF 0x3f3f3f3f
#define LLINF 0x3f3f3f3f3f3f3f3f
#define eps 1e-6
  
using namespace std;
typedef long long LL;
typedef pair<int, int> P;
const int maxn = 500100;
const int mod = 998244353;
int top, a[maxn], b[maxn], c[maxn];

int main()
{
	int n, m, i, j, k, ans = 0, pos;
	scanf("%d %d", &n, &m);
	a[0] = -INF, a[n+1] = INF;
	b[0] = b[n+1] = 1;
	for(i=1;i<=n;i++)scanf("%d", &a[i]);
	for(i=1;i<=m;i++){
		scanf("%d", &j);
		b[j] = 1;
	}
	i = 0;
	while(i<=n){
		for(j=i+1;b[j]!=1;j++);
		if(a[j] < a[i]+(j-i))break;
		top = 0;
		c[top++] = 0;
		for(k=i+1;k<j;k++){
			if(a[k]-a[i]-(k-i) < 0)a[k] = INF;
			else a[k] = a[k] - a[i]-(k-i);
			if(a[k] >= c[top-1])c[top++] = a[k];
			else{
				pos = upper_bound(c, c+top, a[k])-c;
				c[pos] = a[k];
			}
		}
		pos = upper_bound(c, c+top, a[j]-a[i]-(j-i))-c;
		ans += j-i-pos;
		i = j;
	}
	if(i<=n)puts("-1");
	else printf("%d\n", ans);
	return 0;
}
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