【JZOJ6434】【luoguP5665】【CSP-S2019】划分

时间:2019-12-16
本文章向大家介绍【JZOJ6434】【luoguP5665】【CSP-S2019】划分,主要包括【JZOJ6434】【luoguP5665】【CSP-S2019】划分使用实例、应用技巧、基本知识点总结和需要注意事项,具有一定的参考价值,需要的朋友可以参考一下。

description



analysis

  • 首先有一个结论,对于\([1,i]\)区间划分最后一段的和尽量小,答案会更优,具体证明参考毛爷爷的博客

  • \(f[i]\)为满足\([1,i]\)划分最优时、\((f[i],i]\)这段和最小时的最右的端点,最优划分即为从\(n\)开始向\(f\)不断统计

  • 由后一段比前一段大可知\(sum[f[i]]-sum[f[f[i]]]≤sum[i]-sum[f[i]]\),即\(sum[i]≥2sum[f[i]]-sum[f[f[i]]]\)

  • 右边只和\(f[i]\)有关,把右边的记为\(clac(f[i])\),于是可以维护一个\(clac\)值上升的单调队列来求出每一位的\(f\)

  • 具体维护就是,队头不断出队直到\(sum[i]<clac(q[head])\)(因为\(f[i]\)要尽可能大),\(f[i]\)赋为队头

  • \(clac(i)≤clac(q[tail])\)队尾再不断出队(因为\(clac(i)\)更小且\(i\)更靠右),最后\(i\)进队,就完成了队列的维护


code

#include<stdio.h>
#include<string.h>
#include<algorithm>
#define MAXN 40000005
#define ha 1073741824
#define ll long long
#define LL __int128
#define reg register int
#define fo(i,a,b) for (reg i=a;i<=b;++i)
#define fd(i,a,b) for (reg i=a;i>=b;--i)

using namespace std;

int p[100005],l[100005],r[100005];
int f[MAXN],q[MAXN];
ll b[MAXN],sum[MAXN];
int n,type;

inline int read()
{
    ll x=0,f=1;char ch=getchar();
    while (ch<'0' || '9'<ch){if (ch=='-')f=-1;ch=getchar();}
    while ('0'<=ch && ch<='9')x=x*10+ch-'0',ch=getchar();
    return x*f;
}
inline ll clac(ll x){return 2*sum[x]-sum[f[x]];}
inline void print(LL x)
{
    int num[55];num[0]=0;
    while (x)num[++num[0]]=x%10,x/=10;
    while (num[0])printf("%d",num[num[0]--]);
    printf("\n");
}
int main()
{ 
    n=read(),type=read();
    if (type)
    {
        int x=read(),y=read(),z=read();b[1]=read(),b[2]=read();int m=read(),now=1;
        fo(i,1,m)p[i]=read(),l[i]=read(),r[i]=read();
        fo(i,3,n)b[i]=(x*b[i-1]%ha+y*b[i-2]%ha+z)%ha;
        fo(i,1,n)
        {
            if (i>p[now])++now;
            sum[i]=sum[i-1]+(b[i]%(r[now]-l[now]+1))+l[now];
        }
    }
    else {fo(i,1,n)sum[i]=sum[i-1]+read();}
    int head=0,tail=0;
    fo(i,1,n)
    {
        while (head<tail && sum[i]>=clac(q[head+1]))++head;
        f[i]=q[head];
        while (head<tail && clac(i)<=clac(q[tail]))--tail;
        q[++tail]=i;
    }
    LL ans=0;
    for (reg i=n;i;i=f[i])ans+=(LL)(sum[i]-sum[f[i]])*(sum[i]-sum[f[i]]);
    print(ans);
    return 0;
}

原文地址:https://www.cnblogs.com/horizonwd/p/12051647.html