题意：

给你一个序列，你可以改变任意一个数字的大小，代价是改变量

问你使其变成严格单调递增序列的最小代价

思路：

单调不减的最小代价可以用O（n^2）的时间搞出来，而让单调递增转化为单调不减只需要让a[i]-i就可以了

/* ***********************************************
Author        :devil
************************************************ */
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <stack>
#include <map>
#include <string>
#include <cmath>
#include <stdlib.h>
#define LL long long
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define dep(i,a,b) for(int i=a;i>=b;i--)
#define ou(a) printf("%d\n",a)
#define pb push_back
#define mkp make_pair
template<class T>inline void rd(T &x)
{
    char c=getchar();
    x=0;
    while(!isdigit(c))c=getchar();
    while(isdigit(c))
    {
        x=x*10+c-'0';
        c=getchar();
    }
}
#define IN freopen("in.txt","r",stdin);
#define OUT freopen("out.txt","w",stdout);
using namespace std;
const int inf=0x3f3f3f3f;
const int N=3e3+10;
int a[N],b[N],n;
LL dp[N][N];
int main()
{
    #ifndef ONLINE_JUDGE
    //IN
    #endif
    rd(n);
    rep(i,1,n)
    {
        rd(a[i]);
        a[i]-=i;
        b[i]=a[i];
    }
    sort(b+1,b+n+1);
    int bn=unique(b+1,b+n+1)-b-1;
    memset(dp,inf,sizeof(dp));
    rep(i,1,n) rep(j,1,bn) dp[i][j]=(i>1)?min(dp[i][j-1],dp[i-1][j]+abs(a[i]-b[j])):min(dp[i][j-1],1ll*abs(a[i]-b[j]));
    printf("%I64d\n",dp[n][bn]);
    return 0;
}
View Code 
/* ***********************************************
Author        :devil
************************************************ */
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <stack>
#include <map>
#include <string>
#include <cmath>
#include <stdlib.h>
#define LL long long
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define dep(i,a,b) for(int i=a;i>=b;i--)
#define ou(a) printf("%d\n",a)
#define pb push_back
#define mkp make_pair
template<class T>inline void rd(T &x)
{
    char c=getchar();
    x=0;
    while(!isdigit(c))c=getchar();
    while(isdigit(c))
    {
        x=x*10+c-'0';
        c=getchar();
    }
}
#define IN freopen("in.txt","r",stdin);
#define OUT freopen("out.txt","w",stdout);
using namespace std;
const int inf=0x3f3f3f3f;
const int N=3e3+10;
int a[N],b[N],n;
LL dp[N][N];
int main()
{
    #ifndef ONLINE_JUDGE
    //IN
    #endif
    rd(n);
    rep(i,1,n)
    {
        rd(a[i]);
        a[i]-=i;
        b[i]=a[i];
    }
    sort(b+1,b+n+1);
    rep(i,1,n)
    {
        LL mi=dp[i-1][1];
        rep(j,1,n)
        {
            mi=min(mi,dp[i-1][j]);
            dp[i][j]=abs(a[i]-b[j])+mi;
        }
    }
    rep(i,2,n) dp[n][1]=min(dp[n][1],dp[n][i]);
    printf("%I64d\n",dp[n][1]);
    return 0;
}
View Code