luogu P3195 [HNOI2008]玩具装箱TOY

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比较经典的斜率优化...

斜率优化相当于单调队列的升级版

主要针对的是a[i]*b[j]这一项的优化

对于可能决策的点 在坐标平面中可以表示成(b[j],f(j))这种形式

然后就是找一条合法斜率的直线使得截距最小

然后套用单调队列处理 

写的时候可以选择把求斜率变成函数 这样main里面就套路一些

注意最后队列判空的时候因为要访问tail-1所以注意一下不要越界...

Code:

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<vector>
#include<queue>
#define rep(i,a,n) for(int i = a;i <= n;++i)
#define per(i,n,a) for(int i = n;i >= a;--i)
#define inf 2147483647
#define ms(a,b) memset(a,b,sizeof a)
using namespace std;
typedef long long ll;
typedef double D;
ll read() {
    ll as = 0,fu = 1;
    char c = getchar();
    while(c < '0' || c > '9') {
        if(c == '-') fu = -1;
        c = getchar();
    }
    while(c >= '0' && c <= '9') {
        as = as * 10 + c- '0';
        c = getchar();
    }
    return as * fu;
}
//head
const int N = 100005;
int n;
D s[N],a[N],dp[N],L;
D B(int i) {return a[i]+L+1;}
D x(int i) {return B(i);}
D y(int i) {return dp[i] + B(i)*B(i);}
D k(int i,int j) {return (y(j)-y(i))/(x(j)-x(i));}
int q[N];
int main() {
    n = read(),L = (D)read();
    rep(i,1,n) s[i] = s[i-1] + read(),a[i] = s[i] + i;
    int h = 1,t = 1;
    rep(i,1,n) {
        while(h < t && k(q[h],q[h+1]) < 2 * a[i]) h++;
        dp[i] = dp[q[h]] + (a[i] - B(q[h])) * (a[i] - B(q[h]));
        while(h < t && k(i,q[t-1]) < k(q[t-1],q[t])) t--;
        q[++t] = i;
    }
    printf("%lld\n",(ll)dp[n]);
    return 0;
}