[Dynamic Planning] packing toys

Professor descriptionp is going to see the Olympics, but he can't leave his toys, so he decided to transport all the toys to Beijing. He uses his own compress to compress, which can convert any item into a pile and put it in a special one-dimensional container. Professor P has the number 1... N pieces of N toys, I pieces of toys after compression into a one-dimensional length CI. to facilitate sorting, Professor P asked that the number of toys in a one-dimensional container be continuous. At the same time, if a one-dimensional container contains multiple toys, a unit-length padding should be added between the two toys, in the form of placing the I toys into the J toys in a container, the container length will be X = J-I + sigma (CK) I <= k <= j the cost of creating a container is related to the length of the container. According to Professor's research, if the length of the container is X, the cost of making the container is (X-L) ^ 2. L is a constant. Professor P does not care about the number of containers. He can make containers of the specified length, or even exceed L. But he wants the minimum fee. the first line of input is two integers n, L. enter ci.1 <= n <= 434214, 1 <= L, CI <= 10 ^ 7output output minimum cost sample input5 sample output1 in the next n rows

Set the S array as the prefix and make G [I] = I + s [I].

Simple equation: F [I] = max {f [J] + sqr (G [I]-G [J]-l-len )}.
Set J to the current optimal decision, and K to the right of J. through simplification, the slope inequality can be obtained:
F [J]-f [k]
------- + G [J] + G [k] + 2len + 2> = 2G [I]
G [J]-G [k]
Let the left side of the inequality be H (j, k ).

You only need to use a dual-end queue to maintain a decision sequence with an increasing h value (lower convex function.
Accode:

#include <cstdio>#include <cstring>#include <string>#include <cstdlib>#include <algorithm>const char fi[] = "toy.in";const char fo[] = "toy.out";const int maxN = 50010;const int MAX = 0x3f3f3f3f;const int MIN = ~MAX;typedef long long int64;int64 F[maxN], g[maxN], s[maxN], len;int q[maxN], n, f, r;void init_file(){    freopen(fi, "r", stdin);    freopen(fo, "w", stdout);    return;}inline int getint(){    int res = 0; char tmp;    while (!(isdigit(tmp = getchar())));    do res = (res << 3) + (res << 1) + tmp - '0';    while ((isdigit(tmp = getchar())));    return res;}void readdata(){    n = getint(); len = getint();    for (int i = 1; i < n + 1; ++i)    {        (s[i] = getint()) += s[i - 1];        g[i] = i + s[i];    }    return;}#define sqr(x) ((x) * (x))#define check(j, k, i) \(F[j] - F[k] + (g[j] + g[k] + \(len << 1) + 2) * (g[j] - g[k]) \<= (g[i] * (g[j] - g[k]) << 1))#define cmp(j, k, i) \((F[j] - F[k] + (g[j] + g[k] + \(len << 1) + 2) * (g[j] - g[k])) \* (g[k] - g[i]) \<= (F[k] - F[i] + (g[k] + g[i] + \(len << 1) + 2) * (g[k] - g[i])) \* (g[j] - g[k]))void work(){    f = 0, r = 1; //    for (int i = 1; i < n + 1; ++i)    {        while (f < r - 1 && !check(q[f], q[f + 1], i)) ++f;        F[i] = F[q[f]] + sqr(g[i] - g[q[f]] - 1 - len);        while (f < r - 1 && !cmp(q[r - 2], q[r - 1], i)) --r;        q[r++] = i;    }    printf("%I64d\n", F[n]);    return;}int main(){    init_file();    readdata();    work();    return 0;}#undef sqr#undef check#undef cmp