HDU 3507 Print Article（斜率优化）

Print Article

Time Limit: 9000/3000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 15536    Accepted Submission(s): 4813

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

 

题意

给出n个数，可以分成任意连续的段，每段的花费是这段内的数字和的平方，加m（定值），求最小花费。

code

#include<cstdio>
#include<algorithm>

using namespace std;

const int N = 500100;
const int INF = 1e9;
typedef long long LL;
LL s[N],f[N];
int q[N];

int Slope(int j,int k) {
    if (s[j]==s[k]) 
        if (f[j] > f[k]) return -INF;
        else return INF;
    return ((f[j]+s[j]*s[j])-(f[k]+s[k]*s[k]))/(2*s[j]-2*s[k]);
}

int main() {
    int n,m;
    while (~scanf("%d%d",&n,&m)) {
        for (int i=1; i<=n; ++i) {
            scanf("%lld",&s[i]);s[i] += s[i-1];
        }
        int L = 0,R = 0;
        for (int i=1; i<=n; ++i) {
            while (L<R && Slope(q[L+1],q[L])<s[i]) L++;
            int j = q[L];
            f[i] = f[j] + (s[i]-s[j])*(s[i]-s[j])+m;
            while (L<R && Slope(q[R],q[R-1])>Slope(i,q[R])) R--;
            q[++R] = i;
        }
        printf("%lld
",f[n]);
    }
    return 0;
}

参考：

斜率优化

http://www.cnblogs.com/ka200812/archive/2012/08/03/2621345.html

http://www.cnblogs.com/xiaolongchase/archive/2012/02/10/2344769.html