HDU 3507 Print Article

Print Article

Time Limit: 3000ms

Memory Limit: 65536KB

This problem will be judged on HDU. Original ID: 3507
64-bit integer IO format: %I64d      Java class name: Main

 

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

Source

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

 

解题：斜率优化dp。。。

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <climits>
#include <vector>
#include <queue>
#include <cstdlib>
#include <string>
#include <set>
#include <stack>
#define LL long long
#define pii pair<int,int>
#define INF 0x3f3f3f3f
using namespace std;
const int maxn = 500010;
LL sum[maxn],dp[maxn];
int n,m,q[maxn],head,tail;
LL G(int j,int k){
    LL a = dp[j]+sum[j]*sum[j];
    LL b = dp[k]+sum[k]*sum[k];
    return a-b;
}
LL S(int j,int k){
    return (sum[j] - sum[k])*2;
}
int main() {
    int i;
    while(~scanf("%d %d",&n,&m)){
        sum[0] = 0;
        for(i = 1; i <= n; i++){
            scanf("%I64d",sum+i);
            sum[i] += sum[i-1];
        }
        q[0] = dp[0] = head = tail = 0;
        for(i = 1; i <= n; i++){
            while(head < tail && G(q[head+1],q[head]) <= sum[i]*S(q[head+1],q[head])) ++head;
            dp[i] = dp[q[head]] + (sum[i] - sum[q[head]])*(sum[i] - sum[q[head]])+m;
            while(head < tail && G(q[tail-1],q[tail])*S(q[tail],i) >= G(q[tail],i)*S(q[tail-1],q[tail])) --tail;
            q[++tail] = i;
        }
        printf("%I64d
",dp[n]);
    }
    return 0;
}