p2501 [HAOI2006]数字序列

传送门

分析

https://www.luogu.org/blog/FlierKing/solution-p2501

对于第二问的感性理解就是有上下两条线，一些点在上面的线的上面或者下面的线的下面，然后看它们变成哪个线的位置更优

代码

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<cctype>
#include<cmath>
#include<cstdlib>
#include<queue>
#include<ctime>
#include<vector>
#include<set>
#include<map>
#include<stack>
using namespace std;
#define int long long
const int inf = 1e9+7;
int a[100100],d[100100],dp[100100],ddp[100100],sum1[100100],sum2[100100];
vector<int>v[100100];
signed main(){
    int n,m,i,j,k,x,Ans=0;
    scanf("%lld",&n);
    for(i=1;i<=n;i++)scanf("%lld",&x),a[i]=x-i;
    a[++n]=inf;d[0]=a[0]=-inf;
    for(i=1;i<=n;i++)d[i]=inf;
    for(i=1;i<=n;i++){
      int p=upper_bound(d,d+Ans+1,a[i])-d;
      Ans=max(Ans,p);
      dp[i]=p;
      d[p]=a[i];
    }
    v[0].push_back(0);
    for(i=1;i<=n;i++)v[dp[i]].push_back(i);
    printf("%lld\n",n-Ans);
    ddp[0]=0;
    for(i=1;i<=n;i++)ddp[i]=inf;
    for(i=1;i<=n;i++)
      for(j=0;j<v[dp[i]-1].size();j++){
        int to=v[dp[i]-1][j];
        if(to>i)break;
        if(a[to]>a[i])continue;
        for(k=to;k<=i;k++)sum1[k]=abs(a[k]-a[to]),sum2[k]=abs(a[k]-a[i]);
        for(k=to+1;k<=i;k++)sum1[k]+=sum1[k-1],sum2[k]+=sum2[k-1];
        for(k=to;k<i;k++)
          ddp[i]=min(ddp[i],ddp[to]+sum1[k]-sum1[to]+sum2[i]-sum2[k]);
      }
    printf("%lld\n",ddp[n]);
    return 0;
}

转载于:https://www.cnblogs.com/yzxverygood/p/10354437.html