HDU 3507(斜率优化dp-优快云博客

本文深入探讨了一种使用斜率优化的动态规划算法解决选取连续数字最小代价问题的方法。通过详细解释算法流程和代码实现，帮助读者理解如何在特定条件下优化DP过程，提高算法效率。

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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;
}