HDU3507 Print Article(斜率优化DP)

Time Limit: 9000/3000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 16213    Accepted Submission(s): 4992

Problem Description

Zero has an old printer that doesn't work well sometimes. As it is antique, he still like to use it to print articles. But it is too old to work for a long time and it will certainly wear and tear, so Zero use a cost to evaluate this degree.
One day Zero want to print an article which has N words, and each word i has a cost Ci to be printed. Also, Zero know that print k words in one line will cost

M is a const number.
Now Zero want to know the minimum cost in order to arrange the article perfectly.

 

 

Input

There are many test cases. For each test case, There are two numbers N and M in the first line (0 ≤ n ≤ 500000, 0 ≤ M ≤ 1000). Then, there are N numbers in the next 2 to N + 1 lines. Input are terminated by EOF.

 

 

Output

A single number, meaning the mininum cost to print the article.

 

 

Sample Input

5 5 5 9 5 7 5

 

 

Sample Output

230

 

 

Author

Xnozero

 

 

Source

2010 ACM-ICPC Multi-University Training Contest（7）——Host by HIT

 

 

Recommend

zhengfeng   |   We have carefully selected several similar problems for you:  3506 3501 3504 3505 3498 

 

 

比较裸的斜率优化。

以前用的是斜截式推，现在改用求决策单调性的方式推了，

后者虽然计算量大，但是无脑一些

推出来之后维护一个上凸壳

推荐一篇写的非常好的博客

https://www.cnblogs.com/orzzz/p/7885971.html

不知道为啥不能写成除法的形式。。

 

#include<cstdio>
#include<cstring>
using namespace std;
const int MAXN = 1e6 + 10, INF = 1e9 + 10;
int N, M, a[MAXN], s[MAXN], f[MAXN], q[MAXN];
int X(int x) {
    return s[x];
}
int Y(int x) {
    return f[x] + s[x] * s[x];
}
int check(int x, int y, int i) {
    return (Y(y) - Y(x)) <= (X(y) - X(x)) * 2 * s[i];
}
int check2(int xx1, int yy1, int xx2, int yy2) {
    return ((Y(yy1) - Y(xx1)) * (X(yy2) - X(xx2))) >= ((Y(yy2) - Y(xx2)) * (X(yy1) - X(xx1)));
}
/*int slope(int x, int y) {
    return (Y(y) - Y(x)) / (X(y) - X(x));
}*/
int main() {
    //freopen("a.in", "r", stdin);
    while(scanf("%d %d", &N, &M) != EOF) {
        memset(f, 0, sizeof(f)); 
        for(int i = 1; i <= N; i++) scanf("%d", &a[i]), s[i] = s[i - 1] + a[i];
        int h = 0, t = 0;
        for(int i = 1; i <= N; i++) {
        //    if(h < t)    printf("%lf %lf\n", slope(q[h], q[h + 1]), (double)2 * s[i]);
            while(h < t && check(q[h], q[h + 1], i)) h++;
            f[i] = (f[q[h]] + (s[i] - s[q[h]]) * (s[i] - s[q[h]]) + M);
            while(h < t && check2(q[t - 1], q[t], q[t], i)) t--;
            q[++t] = i;
        }
        printf("%d\n", f[N]);
    }
    return 0;
}