Codeforces Round #371 (Div. 2) E 【DP+离散化 LIS 】用最小代价把序列变成严格递增序列

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传送门：codeforces 714 E. Sonya and Problem Wihtout a Legend

描述:

Sonya was unable to think of a story for this problem, so here comes the formal description.

You are given the array containing n positive integers. At one turn you can pick any element and increase or decrease it by 1. The goal is the make the array strictly increasing by making the minimum possible number of operations. You are allowed to change elements in any way, they can become negative or equal to 0.

Input

The first line of the input contains a single integer n (1 ≤ n ≤ 3000) — the length of the array.

Next line contains n integer ai (1 ≤ ai ≤ 109).

Output

Print the minimum number of operation required to make the array strictly increasing.

Examples
input
72 1 5 11 5 9 11
output
9
input
55 4 3 2 1
output
12

Note

In the first sample, the array is going to look as follows:

2 3 5 6 7 9 11

|2 - 2| + |1 - 3| + |5 - 5| + |11 - 6| + |5 - 7| + |9 - 9| + |11 - 11| = 9

And for the second sample:

1 2 3 4 5

|5 - 1| + |4 - 2| + |3 - 3| + |2 - 4| + |1 - 5| = 12

题意：

给一个序列，可以给每一个数加减一个数，代价为他们改变的数的绝对值，那么要求用最小代价把序列变成严格递增的

思路：

有一个非严格递增的版本：POJ3666 题解见Here

而本题的严格如何转化非严格，直接加一行代码，a[i]-=i；

原理

推导过程大致如下：a[i] < a[i+1] —> a[i] <= a[i+1]-1 —> a[i]-i <= a[i+1]-(i+1)—> b[i]<=b[i+1]

代码：

#include <bits/stdc++.h>#define ll __int64using  namespace  std;const int N=3005;int n,a[N],b[N];ll dp[N][N];template<class T> void read(T&num) {    char CH; bool F=false;    for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar());    for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar());    F && (num=-num);}void solve(){  for(int i=1; i<=n; i++){    ll mn=dp[i-1][1];    for(int j=1; j<=n; j++){      mn=min(mn, dp[i-1][j]);      dp[i][j]=abs(a[i]-b[j])+mn;    }  }  ll ans=dp[n][1];  for(int i=2; i<=n; i++){    ans=min(ans, dp[n][i]);  }  printf("%I64d\n",ans);}int  main(){  read(n);  for(int i=1; i<=n; i++){    read(a[i]);    a[i]-=i;    b[i]=a[i];  }  sort(b+1, b+n+1);  solve();  return 0;}

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