Sonya and A Problem without A Legend(DP+优化)

最新推荐文章于 2020-04-05 12:09:39 发布

Twillz

最新推荐文章于 2020-04-05 12:09:39 发布

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给定一个包含正整数的数组，每次操作可以选择一个元素增加或减少1，目标是通过最少的操作次数使数组变得严格递增。输入包含数组长度和数组元素，输出需要的最小操作数。通过动态规划和优化方法可以解决此问题，记录上一层的最小值来减少计算量。

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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.

Example
Input
7
2 1 5 11 5 9 11
Output
9
Input
5
5 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：一次操作对一个数可以+1，-1，求让一个序列变成严格单调递增的需要多少操作；

严格单调递增变为非严格递增的只需要a[i]-i即可
dp[i][j]表示第i个数字变为j最少需要最少操作，则dp[i][j]=min (dp[i-1][k])+abs(a[i]-j)，dp[i-1][k]取的是i-1中最小的，所以只要开个数组记录一下上一层的最小值就可以实现优化了

#include <iostream>
#include <cmath>
#include <cstring>
#include <cstdio>
#include <algorithm>
#define maxn 3005
using namespace std;
typedef long long ll;
const ll inf = 0x3f3f3f3f3f3f;
ll a[maxn];
ll b[maxn];
ll n;
ll dp[maxn][maxn];
ll d[maxn];
int main()
{
    while(scanf("%I64d",&n)==1)
    {

        for(int i=1;i<=n;i++)
        {
            scanf("%I64d",&a[i]);
            a[i]=a[i]-i;
            b[i]=a[i];
        }
        sort(b+1,b+n+1);
        memset(dp,0,sizeof(dp));
        memset(d,inf,sizeof(d));
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=n;j++)
            {
                if(i==1)dp[i][j]=abs(a[i]-b[j]);
                else dp[i][j]=d[j]+abs(a[i]-b[j]);
                d[j]=min(d[j-1],dp[i][j]);
            }
        }
        printf("%I64d\n",d[n]);
    }
    return 0;
}