牛客练习赛72-C brz的序列 (下凸壳，斜率优化)
题面：
思路：
我们可以推理出如下规律：
选择任意(lleq r)，可以使(a_l,dots,a_r)变为首项为(a_l)，尾项为(a_r)的等差数列。
那么本题转化为了选择若干个(a_i)，作为等差数列的首尾相，使总和最小。
为了更好的解决该问题，我们把数(a_i)，转为二维平面中坐标为((i,a_i)) 的点，
那么根据等差数列的性质可以得知，等差数列的公差为数列中两项对应在二维平面上两点的斜率。
通过分析可以发现，将点集维护成下凸壳（凸包的下半部分，见下图）的形式可以使答案最优。
蓝色的点是下凸壳，红色的点是应该从点集中踢出的点。
那么我们用一个栈来维护递增的斜率即可。
代码：
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <bits/stdc++.h>
#define ALL(x) (x).begin(), (x).end()
#define sz(a) int(a.size())
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), ' ', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define chu(x) if(DEBUG_Switch) cout<<"["<<#x<<" "<<(x)<<"]"<<endl
#define du3(a,b,c) scanf("%d %d %d",&(a),&(b),&(c))
#define du2(a,b) scanf("%d %d",&(a),&(b))
#define du1(a) scanf("%d",&(a));
using namespace std;
typedef long long ll;
ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;}
ll lcm(ll a, ll b) {return a / gcd(a, b) * b;}
ll powmod(ll a, ll b, ll MOD) { if (a == 0ll) {return 0ll;} a %= MOD; ll ans = 1; while (b) {if (b & 1) {ans = ans * a % MOD;} a = a * a % MOD; b >>= 1;} return ans;}
ll poww(ll a, ll b) { if (a == 0ll) {return 0ll;} ll ans = 1; while (b) {if (b & 1) {ans = ans * a ;} a = a * a ; b >>= 1;} return ans;}
void Pv(const vector<int> &V) {int Len = sz(V); for (int i = 0; i < Len; ++i) {printf("%d", V[i] ); if (i != Len - 1) {printf(" ");} else {printf("
");}}}
void Pvl(const vector<ll> &V) {int Len = sz(V); for (int i = 0; i < Len; ++i) {printf("%lld", V[i] ); if (i != Len - 1) {printf(" ");} else {printf("
");}}}
inline long long readll() {long long tmp = 0, fh = 1; char c = getchar(); while (c < '0' || c > '9') {if (c == '-') { fh = -1; } c = getchar();} while (c >= '0' && c <= '9') { tmp = tmp * 10 + c - 48, c = getchar(); } return tmp * fh;}
inline int readint() {int tmp = 0, fh = 1; char c = getchar(); while (c < '0' || c > '9') {if (c == '-') { fh = -1; } c = getchar();} while (c >= '0' && c <= '9') { tmp = tmp * 10 + c - 48, c = getchar(); } return tmp * fh;}
void pvarr_int(int *arr, int n, int strat = 1) {if (strat == 0) {n--;} repd(i, strat, n) {printf("%d%c", arr[i], i == n ? '
' : ' ');}}
void pvarr_LL(ll *arr, int n, int strat = 1) {if (strat == 0) {n--;} repd(i, strat, n) {printf("%lld%c", arr[i], i == n ? '
' : ' ');}}
const int maxn = 1000010;
const int inf = 0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
#define DEBUG_Switch 0
int n;
ll a[maxn];
pll st[maxn];
int main()
{
#if DEBUG_Switch
freopen("D:\code\input.txt", "r", stdin);
#endif
//freopen("D:\code\output.txt","w",stdout);
n = readint();
repd(i, 1, n) {
a[i] = readint();
}
if (n == 1) {
cout << fixed << setprecision(10) << a[1] << endl;
return 0;
}
int l = 0;
int r = 0;
repd(i, 1, n)
{
while (r - l > 1 && ( a[i] - st[r].first) * (i - st[r - 1].second ) < (a[i] - st[r - 1].first ) * (i - st[r].second ))
{
r--;
}
st[++r] = mp(a[i], i);
}
ll ans = 0ll;
ll temp = 0ll;
repd(i, l + 2, r)
{
ans += (st[i].se - st[i - 1].se + 1) * (st[i].fi + st[i - 1].fi);
if (i >= l + 2 && i <= r - 1)
temp += st[i].fi;
}
long double out = 0.5 * ans;
out -= temp;
cout << fixed << setprecision(10) << out << endl;
return 0;
}