题目来源:传送门
题意把序列转化为严格递增,问最小操作数。与poj3666类似,唯一不同于是严格递增是非严格递增。
而严格递增可以转为为非严格递增。
推导:要让序列a[i]<a[i+1]则满足a[i]<=a[i+1]-1,由此可知a[i]-i<=a[i+1]-(i+1),此序列为非严格递增,最终严格递增序列每个a[i]变为a[i]-i,由最初a[i]->严格递增a[i]+x,x为增减值,(a[i]-i)-x为初始构建非严格递增的序列,将(a[i]-i)看做整体,由此可知a[i]-i后直接按照非严格递增序列的做法推导即可。
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<vector>
#include<map>
#include<algorithm>
#include<queue>
#define MAX_len 50100*4
using namespace std;
typedef long long ll;
ll a[3010];
ll dp[3010][3010];
ll b[3010];
main()
{
int n,i,j;
scanf("%d",&n);
ll ans=999999999999999;
for(i=1;i<=n;i++)
{
cin>>a[i];
b[i]=a[i]-i;
a[i]-=i;
}
sort(b+1,b+n+1);
for(i=1;i<=n;i++)
{
ll MIN=99999999999999;
for(j=1;j<=n;j++)
{
MIN=min(dp[i-1][j],MIN);
dp[i][j]=abs(b[j]-a[i])+MIN;
}
}
for(i=1;i<=n;i++)
{
ans=min(ans,dp[n][i]);
}
cout<<ans;
return 0;
}