UESTC

他们oj挂掉啦， 我先保存一下代码。。。

直接dp复杂度， n * 100 * 100， 我们可以将前一个人的信息丢进单调队列中去，可以优化成n * 100;

#include<bits/stdc++.h>
#define LL long long
#define fi first
#define se second
#define mk make_pair
#define pii pair<int,int>
#define piii pair<int, pair<int,int> >

using namespace std;

const int N = 5e4 + 7;
const int M = 10 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;
const double eps = 1e-6;

int n, c, head1, head2, rear1, rear2;
int sk1[N], sk2[N], a[N], dp[N][101];

void init() {
    for(int i = 1; i <= n; i++) {
        for(int j = 0; j <= 100; j++) {
            dp[i][j] = inf;
        }
    }
    head1 = head2 = 0;
    rear1 = rear2 = 0;
}
int main() {
    while(scanf("%d%d", &n, &c) != EOF) {

        init();
        for(int i = 1; i <= n; i++) {
            scanf("%d", &a[i]);
        }

        for(int j = a[1]; j <= 100; j++) {
            dp[1][j] = (j - a[1]) * (j - a[1]);
        }

        for(int i = 2; i <= n; i++) {

            rear1 = rear2 = 0;
            head1 = head2 = 0;

            for(int j = 0; j <= 100; j++) {

                while(head1 < rear1 && dp[i - 1][j] + c * j < dp[i - 1][sk1[rear1 - 1]] + c * sk1[rear1 - 1]) rear1--;
                sk1[rear1++] = j;
            }

            for(int j = 100; j >= 0; j--) {

                while(head2 < rear2 && dp[i - 1][j] - c * j < dp[i - 1][sk2[rear2 - 1]] - c * sk2[rear2 - 1]) rear2--;
                sk2[rear2++] = j;
            }

            for(int j = a[i]; j <= 100; j++) {
                while(head1 < rear1 && sk1[head1] < j) head1++;
                if(head1 < rear1) dp[i][j] = dp[i - 1][sk1[head1]] + c * sk1[head1] + (j - a[i]) * (j - a[i]) - c * j;
            }

            for(int j = 100; j >= a[i]; j--) {
                while(head2 < rear2 && sk2[head2] > j) head2++;
                if(head2 < rear2) dp[i][j] = min(dp[i][j], dp[i - 1][sk2[head2]] - c * sk2[head2] + (j - a[i]) * (j - a[i]) + c * j);
            }
        }

        int ans = inf;
        for(int j = a[n]; j <= 100; j++) {
            ans = min(ans, dp[n][j]);
        }
        printf("%d
", ans);
    }
    return 0;
}
/*
*/