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題意與思路：點擊打開鏈接

 

AC code

 

#include<cstdio>
#include<iostream>
#include<cstring>
using namespace std;

const int MAXN = 500010;
int dp[MAXN];
int q[MAXN];//隊列
int sum[MAXN];

int head, tail, n, m;// dp[i]= min{ dp[j]+M+(sum[i]-sum[j])^2 };

int getDP(int i, int j)
{
	return dp[j] + m + (sum[i] - sum[j])*(sum[i] - sum[j]);
}

int getUP(int j, int k) //yj-yk部分
{
	return dp[j] + sum[j] * sum[j] - (dp[k] + sum[k] * sum[k]);
}

int getDOWN(int j, int  k)
{
	return 2 * (sum[j] - sum[k]);
}

int main()
{
	while (scanf("%d%d", &n, &m) == 2)
	{
	    sum[0] = dp[0] = 0;
		for (int i = 1; i <= n; i++)    scanf("%d", &sum[i]);
		for (int i = 1; i <= n; i++)    sum[i] += sum[i - 1];
		head = tail = 0;
		q[tail++] = 0;
		for (int i = 1; i <= n; i++)
		{
			while (head + 1 < tail &&  getUP(q[head + 1], q[head]) <= sum[i] * getDOWN(q[head + 1], q[head]))
				head++;
			dp[i] = getDP(i, q[head]);
			while (head + 1 < tail && getUP(i, q[tail - 1])*getDOWN(q[tail - 1], q[tail - 2]) <= getUP(q[tail - 1], q[tail - 2])*getDOWN(i, q[tail - 1]))
				tail--;
			q[tail++] = i;
		}
		printf("%d\n", dp[n]);
	}
	return 0;
}