HDU 3507(斜率优化dp

题目：每次选取连续的若干数字的代价

要求选取虽有数字的最小代价．

思路：基础斜率ｄｐ题，题解见http://www.cnblogs.com/kuangbin/archive/2012/08/26/2657650.html

/*
* @author:  Cwind
*/
#pragma comment(linker, "/STACK:102400000,102400000")
#include <iostream>
#include <map>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <vector>
#include <queue>
#include <stack>
#include <functional>
#include <set>
#include <cmath>
using namespace std;
#define IOS std::ios::sync_with_stdio (false);std::cin.tie(0)
#define pb push_back
#define PB pop_back
#define bk back()
#define fs first
#define se second
#define sq(x) (x)*(x)
#define eps (1e-6)
#define IINF (1<<29)
#define LINF (1ll<<59)
#define INF (1000000000)
#define FINF (1e3)
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
typedef pair<ll,ll> P;

const int maxn=5e5+3000;
int n,m;
int a[maxn];
int Q[maxn];
int sum[maxn];
int dp[maxn];
int front,back;
int get1(int a,int b){
    return dp[b]+sq(sum[b])-dp[a]-sq(sum[a]);
}
int get2(int a,int b){
    return 2*(sum[b]-sum[a]);
}
int main(){
    freopen("/home/files/CppFiles/in","r",stdin);
    //freopen("test.in","r",stdin);
    //freopen("test.out","w",stdout);
    while(cin>>n>>m){
        front=back=0;
        for(int i=1;i<=n;i++){
            scanf("%d",&a[i]);
            sum[i]=sum[i-1]+a[i];
        }
        Q[back++]=0;
        for(int i=1;i<=n;i++){
            while(front+1<back&&get1(Q[front],Q[front+1])<=sum[i]*get2(Q[front],Q[front+1])){
                 front++;
            }
            dp[i]=dp[Q[front]]+sq(sum[i]-sum[Q[front]])+m;
            while(front+1<back&&get1(Q[back-2],Q[back-1])*get2(Q[back-1],i)>=get1(Q[back-1],i)*get2(Q[back-2],Q[back-1])){
                back--;
            }
            Q[back++]=i;
        }
        printf("%d\n",dp[n]);
    }
    return 0;
}