UESTC 594 我要长高 (单调队列优化DP)

题意

给出一个序列，没两个之间差值乘C就是亏损的钱数，可以给一些数列以一定的代价变大，不能减小，求增加方案令最后的损失最小。

思路

令dp[i][j]表示第i个人身高为j时的当前损失。
我们能得到 dp[i][j]=min(dp[i−1][k]+abs(j−k)∗C+(x[i]−j)∗(x[i]−j))
我们需要枚举的有当前的人i，当前身高j前一个人身高k，枚举前一个人身高可以用单调队列优化掉，总复杂度 O(nh) 。

代码

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
#define LL long long
#define lowbit(x) ((x)&(-x))
#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1|1
#define MP(a, b) make_pair(a, b)
const int INF = 0x3f3f3f3f;
const int maxn = 1e5 + 7;
const double eps = 1e-8;
const int MOD = 1000000009;
const double PI = acos(-1.0);

int dp[2][110];
int n, c;
int q[110];
int head, tail, cur;

int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    while (scanf("%d%d", &n, &c) != EOF)
    {
        int x;
        scanf("%d", &x);
        cur = 0;
        for (int i = 0; i < x; i++)
            dp[cur][i] = INF;
        for (int i = x; i <= 100; i++)
            dp[cur][i] = (x - i) * (x - i);
        for (int i = 1; i < n; i++)
        {
            scanf("%d", &x);
            cur ^= 1;
            head = tail = 0;
            for (int j = 0; j <= 100; j++)
            {
                int num = dp[cur^1][j] - c * j;
                while (head < tail && q[tail-1] > num)
                    tail--;
                q[tail++] = num;
                if (j < x)
                    dp[cur][j] = INF;
                else
                    dp[cur][j] = q[head] + j * c + (x - j) * (x - j);
            }
            head = tail = 0;
            for (int j = 100; j >= 0; j--)
            {
                int num = dp[cur^1][j] + j * c;
                while (head < tail && q[tail-1] > num)
                    tail--;
                q[tail++] = num;
                if (j >= x)
                    dp[cur][j] = min(dp[cur][j], q[head] - j * c + (x - j) * (x - j));
            }
        }
        int ans = INF;
        for (int i = 0; i <= 100; i++)
            ans = min(ans, dp[cur][i]);
        printf("%d\n", ans);
    }
    return 0;
}