hud 3507，Print Article，斜率优化dp

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#算法 #动态规划

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这篇文章介绍了一道ACM-ICPC编程题，涉及如何使用斜率优化的动态规划方法来计算在给定成本限制下，将一篇文章按照最优顺序打印所需的最小成本。通过计算单词之间的成本差值，确定每行的最小成本组合。

Problem - 3507 (hdu.edu.cn)

Time Limit: 9000/3000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 26705 Accepted Submission(s): 8024

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

Sample Output

230

Author

Xnozero

Source

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

解析：斜率优化dp

HDU3507 Print Article —— 斜率优化DP - h_z_cong - 博客园 (cnblogs.com)

#include<iostream>
#include<string>
#include<cstring>
#include<cmath>
#include<ctime>
#include<algorithm>
#include<utility>
#include<stack>
#include<queue>
#include<vector>
#include<set>
#include<math.h>
#include<map>

using namespace std;
typedef long long LL;
const int N = 5e5 + 5;
int n, m;
LL c[N], f[N],q[N];

LL getup(int j, int k) {
	return f[j] + c[j] * c[j] - (f[k] + c[k] * c[k]);
}

LL getdown(int j, int k) {
	return 2 * (c[j] - c[k]);
}

int main() {
	while (scanf("%d%d", &n, &m) != EOF) {
		for (int i = 1; i <= n; i++) {
			scanf("%d", &c[i]);
			c[i] += c[i - 1];
		}
		f[0] = 0;
		int hh = 0, tt = 0;
		q[0] = 0;
		for (int i = 1; i <= n; i++) {
			while (hh < tt && getup(q[hh + 1], q[hh]) < getdown(q[hh + 1],q[ hh]) * c[i])hh++;
			int j = q[hh];
			f[i] = f[j] + (c[i] - c[j]) * (c[i] - c[j]) + m;
			while (hh < tt && getup(i, q[tt]) * getdown(q[tt], q[tt - 1]) <= getup(q[tt], q[tt - 1]) * getdown(i, q[tt]))tt--;
			q[++tt] = i;
		}
		printf("%d\n", f[n]);
	}
	return 0;
}