BZOJ1911 [APIO2010] 特别行动队

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

Description

Input

Output

斜率优化DP

教主的题解：http://www.cnblogs.com/JSZX11556/p/4811459.html

我也懒得写下去了……

1.10晚 补题解：令 s[i] 为 x[i] 的前缀和，f[i] = min{ f[j] + A * ( s[i] - s[j] ) ^ 2 + B * ( s[i] - s[j] ) + C }， j < i

若 j 优于 k，j > k，则有 f[j] - f[k] + A * ( s[j] ^ 2 - s[k] ^ 2 ) >= ( 2 * A * s[i] + B ) *  ( s[j] - s[k] )

将 s[j] - s[k] 移到左边，得斜率单调递减，用单调队列维护一个下凸函数即可。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#define rep(i,l,r) for(int i=l; i<=r; i++)
#define clr(x,y) memset(x,y,sizeof(x))
using namespace std;
typedef long long ll;
const int maxn = 1000010;
ll n,A,B,C,head,tail,f[maxn],q[maxn],s[maxn];
inline ll read(){
    ll ans = 0, f = 1;
    char c = getchar();
    for(; !isdigit(c); c = getchar())
    if (c == '-') f = -1;
    for(; isdigit(c); c = getchar())
    ans = ans * 10 + c - '0';
    return ans * f;
}
inline ll calc(int x){
    return f[x] + A * s[x] * s[x];
}
inline ll get(int i,int j){
    return f[j] + A * (s[i] - s[j]) * (s[i] - s[j]) + B * (s[i] - s[j]) + C;
}
int main(){
    n = read(); A = read(); B = read(); C = read();
    rep(i,1,n) s[i] = s[i-1] + read();
    head = tail = f[0] = q[0] = 0;
    rep(i,1,n){
        while (head < tail && calc(q[head+1]) - calc(q[head]) >= (2 * A * s[i] + B) * (s[q[head+1]] - s[q[head]])) head++;
        f[i] = get(i,q[head]);
        while (head < tail && (calc(q[tail]) - calc(q[tail-1])) * (s[i] - s[q[tail]]) <= (calc(i) - calc(q[tail])) * (s[q[tail]] - s[q[tail-1]])) tail--;
        q[++tail] = i;
    }
    printf("%lld\n",f[n]);
    return 0;
}