【斜率优化】洛谷_3195 [HNOI2008]玩具装箱TOY

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题意

给出 $N$个物体，每个物体有一个 $C_i$，每次选择压缩第 $i$个物体到第 $j$个物体的代价是 $x=j-i+\sum^{j}_{k=i}C_k$，还要加上 $(x-l)^2$，求出一种压缩方法求出最小的代价。

思路

入门博客

代码

#include<cstdio>

int n, l;
long long s[50001], q[50001], f[50001];

long long x(int i) {
    return s[i] + l;
}

long long y(int i) {
    return f[i] + (s[i] + l) * (s[i] + l);
}

int main() {
    scanf("%d %d", &n, &l);
    l++;
    for (int i = 1; i <= n; i++) {
        scanf("%d", &s[i]);
        s[i] += s[i - 1];
    }
    for (int i = 1; i <= n; i++)
        s[i] += i;
    q[1] = 0;
    int head = 1, tail = 1;
    for (int i = 1; i <= n; i++) {
        while (head < tail && y(q[head + 1]) - y(q[head]) <= 2 * s[i] * (x(q[head + 1]) - x(q[head]))) head++;
        f[i] = f[q[head]] + (s[i] - s[q[head]] - l) * (s[i] - s[q[head]] - l);
        while (head < tail && (y(i) - y(q[tail])) * (x(q[tail]) - x(q[tail - 1])) <= (y(q[tail]) - y(q[tail - 1])) * (x(i) - x(q[tail]))) tail--;
        q[++tail] = i;
    }
    printf("%lld", f[n]);
}