hdu3507_斜率dp

Posted 2020-08-09 _lm

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=3507

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Time Limit: 9000/3000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 9699    Accepted Submission(s): 3066

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

推荐博客：http://blog.csdn.net/azheng51714/article/details/8214165

 

 1 //dp[i]=dp[j]+M+(sum[i]-sum[j])^2
 2 //设k<j<i, j比k决策好
 3 //dp[j]+M+(sum[i]-sum[j])^2<dp[k]+M+(sum[i]-sum[k])^2
 4 //(dp[j]+num[j]^2-(dp[k]+num[k]^2))/(2*(num[j]-num[k]))<sum[i]
 5 //dp[j]+num[j]^2-(dp[k]+num[k]^2))  GetUp()
 6 //2*(num[j]-num[k])  GetDown()
 7 #include <algorithm>
 8 #include <iostream>
 9 #include <cstring>
10 #include <cstdlib>
11 #include <cstdio>
12 #include <vector>
13 #include <ctime>
14 #include <queue>
15 #include <list>
16 #include <set>
17 #include <map>
18 using namespace std;
19 #define INF 0x3f3f3f3f
20 typedef long long LL;
21 
22 int dp[500010], n, m, sum[500010], q[500010];
23 int GetDp(int i, int j)
24 {
25     return dp[j] + m + (sum[i]-sum[j])*(sum[i]-sum[j]);
26 }
27 int GetUp(int j, int k)//yj-yk的部分
28 {
29     return dp[j] + sum[j]*sum[j] - (dp[k]+sum[k]*sum[k]);
30 }
31 int GetDown(int j, int k)//xj-xk的部分
32 {
33     return 2 * (sum[j] - sum[k]);
34 }
35 int main()
36 {
37     while(~scanf("%d %d", &n, &m))
38     {
39         sum[0] = dp[0] = 0;
40         for(int i = 1; i <= n; i++){
41             scanf("%d", &sum[i]);
42             sum[i] += sum[i-1];
43         }
44         int head = 0, tail = 0;
45         q[tail++] = 0;
46         for(int i = 1; i <=n; i++)
47         {
48             while(head+1<tail && GetUp(q[head+1],q[head])<=sum[i]*GetDown(q[head+1],q[head]))
49                 head++;
50             dp[i] = GetDp(i, q[head]);
51             while(head+1<tail && GetUp(i, q[tail-1])*GetDown(q[tail-1],q[tail-2])<=GetUp(q[tail-1],q[tail-2])*GetDown(i,q[tail-1]))
52                 tail--;
53             q[tail++] = i;
54         }
55         printf("%d\n", dp[n]);
56     }
57     return 0;
58 }

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