[省选前题目整理][BZOJ 1911][APIO 2010]特别行动队(斜率优化DP)

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题目链接

http://www.lydsy.com/JudgeOnline/problem.php?id=1911

思路

这里写图片描述
注：单调递减的队列里，相邻两个点的连线的斜率是单调递增的，即维护一个下凸壳

代码

#include <iostream>#include <stdio.h>#include <stdlib.h>#include <string.h>#include <algorithm>#define MAXN 1200000using namespace std;typedef long long int LL;LL f[MAXN],sum[MAXN],q[MAXN],a,b,c,n;double getSlope(int x,int y) //求(x,f[x])和(y,f[y])之间的斜率{    return (double)((f[x]+a*sum[x]*sum[x]-b*sum[x])-(f[y]+a*sum[y]*sum[y]-b*sum[y]))/(2*a*(sum[x]-sum[y]));}int main(){    scanf("%lld%lld%lld%lld",&n,&a,&b,&c);    for(int i=1;i<=n;i++)    {        LL x;        scanf("%lld",&x);        sum[i]=sum[i-1]+x;    }    f[0]=0;    int h=1,t=1;    q[h]=0;    for(int i=1;i<=n;i++)    {        while(h<t&&getSlope(q[h],q[h+1])<=sum[i]) h++;        f[i]=f[q[h]]+a*(sum[i]-sum[q[h]])*(sum[i]-sum[q[h]])+b*(sum[i]-sum[q[h]])+c;        while(h<t&&getSlope(q[t-1],q[t])>getSlope(q[t],i)) t--;        q[++t]=i; //!!!!    }    printf("%lld\n",f[n]);    return 0;}

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