bzoj1911 特别行动队

Description

Input

Output

Sample Input

4
-1 10 -20
2 2 3 4

Sample Output

9
 
 
 
斜率优化 推式子
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cstdio>
#include<cmath>
using namespace std;
const int maxn=1e6+10;
long long n,a,b,c,sum[maxn],yy[maxn],zz[maxn],l=1,r=0,dp[maxn];
 
long long aa,fl;char cc;
long long read() {
    aa=0;cc=getchar();fl=1;
    while((cc<'0'||cc>'9')&&cc!='-') cc=getchar();
    if(cc=='-') fl=-1,cc=getchar();
    while(cc>='0'&&cc<='9') aa=aa*10+cc-'0',cc=getchar();
    return aa*fl;
}
 
bool ok(long long x,long long y,long long z) {
    return (yy[y]-yy[x])*(sum[z-1]-sum[y-1])<(yy[z]-yy[y])*(sum[y-1]-sum[x-1]);
}
 
int main() {
    n=read();a=read();b=read();c=read();
    for(int i=1;i<=n;++i) sum[i]=read(),sum[i]+=sum[i-1];
    for(int i=1;i<=n;++i) {
        yy[i]=a*sum[i-1]*sum[i-1]-b*sum[i-1]+dp[i-1];
        while(l<r&&ok(zz[r-1],zz[r],i)) r--;
        zz[++r]=i;
        while(l<r&&-2*a*sum[i]*sum[zz[l+1]-1]+yy[zz[l+1]]>-2*a*sum[i]*sum[zz[l]-1]+yy[zz[l]]) l++;
        dp[i]=-2*a*sum[i]*sum[zz[l]-1]+yy[zz[l]]+a*sum[i]*sum[i]+b*sum[i]+c;
    }
    printf("%lld",dp[n]);
     
    return 0;
}

  

弱者就是会被欺负呀
posted @ 2017-08-30 14:50  shixinyi  阅读(130)  评论(0编辑  收藏  举报