洛谷P3515 [POI2011]Lightning Conductor（决策单调性）

题意

已知一个长度为n的序列a1,a2,...,an。

对于每个1<=i<=n，找到最小的非负整数p满足 对于任意的j, aj < = ai + p - sqrt(abs(i-j))

题解

决策单调性是个好东西

等学会了再滚回来填坑

//minamoto
#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;
#define getc() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++)
char buf[1<<21],*p1=buf,*p2=buf;
template<class T>inline bool cmax(T&a,const T&b){return a<b?a=b,1:0;}
inline int read(){
    #define num ch-'0'
    char ch;bool flag=0;int res;
    while(!isdigit(ch=getc()))
    (ch=='-')&&(flag=true);
    for(res=num;isdigit(ch=getc());res=res*10+num);
    (flag)&&(res=-res);
    #undef num
    return res;
}
char sr[1<<21],z[20];int C=-1,Z;
inline void Ot(){fwrite(sr,1,C+1,stdout),C=-1;}
inline void print(int x){
    if(C>1<<20)Ot();if(x<0)sr[++C]=45,x=-x;
    while(z[++Z]=x%10+48,x/=10);
    while(sr[++C]=z[Z],--Z);sr[++C]='
';
}
const int N=5e5+5;
int n,q[N],k[N],a[N];
double p[N];
inline double calc(int i,int j){return a[j]+sqrt(i-j);}
inline int bound(int x,int y){
    int l=1,r=n,mid,res=r+1;
    while(l<=r){
        mid=l+r>>1;
        if(calc(mid,x)<=calc(mid,y)) res=mid,r=mid-1;
        else l=mid+1;
    }
    return res;
}
void work(){
    for(int i=1,h=1,t=0;i<=n;++i){
        while(h<t&&k[t-1]>=bound(q[t],i)) --t;
        k[t]=bound(q[t],i),q[++t]=i;
        while(h<t&&k[h]<=i) ++h;
        cmax(p[i],calc(i,q[h]));
    }
}
int main(){
    //freopen("testdata.in","r",stdin);
    n=read();
    for(int i=1;i<=n;++i) a[i]=read();
    work();
    for(int i=1;i<=n+1-i;++i)
    swap(a[i],a[n-i+1]),swap(p[i],p[n-i+1]);
    work();
    for(int i=n;i;--i) print(ceil(p[i])-a[i]);
    Ot();
    return 0;
}