Covered Walkway

Time Limit: 30000/10000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1496    Accepted Submission(s): 602

Problem Description

Your university wants to build a new walkway, and they want at least part of it to be covered. There are certain points which must be covered. It doesn’t matter if other points along the walkway are covered or not. 
The building contractor has an interesting pricing scheme. To cover the walkway from a point at x to a point at y, they will charge c+(x-y)², where c is a constant. Note that it is possible for x=y. If so, then the contractor would simply charge c. 
Given the points along the walkway and the constant c, what is the minimum cost to cover the walkway?

Input

There will be several test cases in the input. Each test case will begin with a line with two integers, n (1≤n≤1,000,000) and c (1≤c≤10⁹), where n is the number of points which must be covered, and c is the contractor’s constant. Each of the following n lines will contain a single integer, representing a point along the walkway that must be covered. The points will be in order, from smallest to largest. All of the points will be in the range from 1 to 10⁹, inclusive. The input will end with a line with two 0s.

Output

For each test case, output a single integer, representing the minimum cost to cover all of the specified points. Output each integer on its own line, with no spaces, and do not print any blank lines between answers. All possible inputs yield answers which will fit in a signed 64-bit integer.

Sample Input

10 5000

1
23
45
67
101
124
560
789
990
1019

0 0

Sample Output

30726

Source

The University of Chicago Invitational Programming Contest 2012

题意：

有n个点需要被覆盖，覆盖第j到第i之间的点的花费是c+(x[i]-x[j])^2，问把所有的点都覆盖的最小花费。

输入n,c

输入n个点x[1...n]

当输入0 0时结束

代码：

//有状态转移方程dp[i]=min(dp[j]+C+(a[i]-a[j+1])*(a[i]-a[j+1])),数据是1e6的两重循环必然不行
//设k<j<i,当到达i点时如果从j点转移到i比从k点转移到i更优则有:dp[j]+C+(a[i]-a[j+1])^2<dp[k]+C+(a[i]-a[k+1])^2
//展开得:(dp[j]-dp[k]+a[j+1]^2-a[k+1]^2)/2*(a[j+1]-a[k+1])<a[i].其中a[i]常量(实现时是一重循环枚举i点)，
//我们设yj=dp[j]+a[j+1]^2,xj=a[j+1] =>(yj-yk)/2*(xj-xk)<a[i].左边是计算斜率的式子。我们用一个单调队列
//来存储能够转移到i点状态的点并且队头是转移到i点状态的最优的解，每次要保持队头是最优解就要根据
//(yj-yk)/2*(xj-xk)<a[i]用队头去和队列中第二个去比较(如果队头不优于第二个就要删去队头元素)。
//设g[i,j]表示直线j-i的斜率，如果有g[k,j]>g[i,j]那么j点永远不可能是i的最优解，因为：
//我们假设g[i,j]<a[i]，那么就是说i点要比j点优，排除j点。如果g[i,j]>=a[i]，那么j点此时是比i点要更优，
//但是同时g[j,k]>g[i,j]>sum[i]。这说明还有k点会比j点更优，同样排除j点。排除多余的点，这便是一种优化！
//其实就是维护一个斜率递增的（下凸上凹）的图形。因此要把i点加入队列之前先判断是否能够维护斜率递增如果
//不能就把队列最后一个元素删掉直到是斜率递增的然后加入i点。
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
typedef long long ll;
const int maxn=1000009;
int n,m,que[maxn];
ll dp[maxn],a[maxn];
ll getdp(int i,int j){
    return dp[j]+m+(a[i]-a[j+1])*(a[i]-a[j+1]);
}
ll getup(int j,int k){
    return dp[j]-dp[k]+a[j+1]*a[j+1]-a[k+1]*a[k+1];
}
ll getlow(int j,int k){
    return 2*(a[j+1]-a[k+1]);
}
int main()
{
    while(scanf("%d%d",&n,&m)&&(n+m)){
        for(int i=1;i<=n;i++) scanf("%lld",&a[i]);
        int head=0,tail=0;
        dp[0]=0;
        que[tail++]=0;
        for(int i=1;i<=n;i++){
            while(head+1<tail&&getup(que[head+1],que[head])<=a[i]*getlow(que[head+1],que[head]))
                head++;
            dp[i]=getdp(i,que[head]);
            while(head+1<tail&&getup(que[tail-1],que[tail-2])*getlow(i,que[tail-1])>=getup(i,que[tail-1])*getlow(que[tail-1],que[tail-2]))
                tail--;
            que[tail++]=i;
        }
        printf("%lld\n",dp[n]);
    }
    return 0;
}