鏂滅巼浼樺寲

Posted 2022-08-19

鏍囩锛?a href='http://www.mamicode.com/so/1/eve' title='eve'>eve   closed   cli   浜ょ偣   lld   mutable   end   style   amp   

鍔熻兘:涓€涓竾鑳界殑鏂滅巼浼樺寲妯℃澘 鍙互瑙ｅ喅妯潗鏍囦笉鍗曡皟 鏌ヨ鍧愭爣涓嶅崟璋冪殑闂

鏂滅巼浼樺寲闂瑙ｅ喅鏂规硶:
鏂滅巼浼樺寲闂鏄綋dp寮忕被浼?dp_i = dp_j + a_i * b_j$褰㈠紡鏃舵棤娉曞乏鍙冲垎绂籭涓巎鏃剁殑涓€绉嶄紭鍖栧鏉傚害鐨勬柟娉曘€?br />
鎴戜滑閫氳繃鍙樻崲寰楀埌绫讳技$y = k * x + b$褰㈠紡,鍏朵腑$x$鍜?y$鏄彧鍏充簬$i$鐨勯」,$k$鍜?b$鏄彧鍏充簬$j$鐨勯」銆傚悓鏃?dp_i$鍦?y$涓?鎴戜滑甯屾湜閫夋嫨鍚堥€傜殑$k$涓?b$浣垮緱$y$鏈€澶ф垨鏈€灏?浠庤€屼娇dp[i]鏈€澶ф垨鏈€灏忋€?br />
浜庢槸鎴戜滑闇€瑕佺淮鎶や竴涓洿绾块泦鍚?(k,b)$,褰撻渶瑕?dp_i$鏈€澶ф椂鐩寸嚎闆嗗悎褰㈡垚涓嬪嚫鐨勫嚫鍖?鍙嶄箣鍒欏舰鎴愪笂鍑哥殑鍑稿寘,姣忔鏌ヨ$query(x)$鍗冲彲寰楀嚭缁撴灉,鍔犲叆鐩寸嚎鍒欐槸$add(k,b)$,澶嶆潅搴?O(nlogn)$銆?/p>

妯℃澘:

namespace 
struct Line 
mutable ll k, m, p;
bool f; // 瀛樺湪鏂滅巼鍚?/span>
Line() 
Line(ll _k, ll _m, ll _p, bool _f) : k(_k), m(_m), p(_p), f(_f)  
bool friend operator < (const Line &a, const Line &b) 
return (a.f && b.f) ? a.k < b.k : a.p < b.p;

;
struct LineContainer : multiset<Line> 
//    LineContainer() 
const ll inf = LLONG_MAX;
ll div(ll a, ll b)  //姹備氦鐐?/span>
return a / b - (a ^ b < 0 && a % b); 

ld div(ld a, ld b) 
return a / b;

bool Intersect(iterator x, iterator y) 
if(y == end()) 
x -> p = inf;
return false;

if(x -> k == y -> k) x -> p = x -> m > y -> m ? inf : -inf;
else x -> p = div(y -> m - x -> m, x -> k - y -> k);
return x -> p >= y -> p;

void add(ll k, ll m) 
multiset<Line> :: iterator z = insert(Line(k, m, 0, 1)), y = z++, x = y;
while(Intersect(y, z)) z = erase(z);
if(x != begin() && Intersect(--x, y)) Intersect(x, y = erase(y));
while((y = x) != begin() && (--x) -> p >= y -> p) Intersect(x, erase(y));

ll query(ll x) 
//    assert(!empty());
multiset<Line> :: iterator L = lower_bound(Line(0, 0, x, 0));
return L -> k * x + L -> m;

;

View Code

渚嬮:
Codeforces 631E
澶ф剰:浣垮緱$\sum_i=1^na_i*i$鏈€澶?鍙互灏嗕竴涓?a_i$鎻掑叆鍒颁换浣曚竴涓綅缃?br />棰樿В:闇€瑕佷袱娆?dp$,娌℃湁宸埆,浠呰鏄庣涓€閬?dp$銆?/p>

鏋氫妇浣嶇疆$i$,鑰冭檻绉诲埌浣嶇疆$j$鍓嶉潰涓?$j \leq i$銆?ans = max(tot + a_i*j-a_i*i+sum_i-1-sum_j-1)$銆傚寲鎴?y = k * x + b$褰㈠紡锛屽緱鍑?ans-tot-sum_i-1+a_i*i=a_i*j-sum_j-1$

鎵€浠?br />$k = j$
$b = -sum_j-1$
$x = a_i$
$y = ans - tot - sum_i-1 + a_i * i$
缁存姢涓嬪嚫澹筹紝鍏蜂綋瑙佷唬鐮併€傜瓟妗堝嵆鏄?query(a_i)+tot+sum_i-1-a_i*i$銆?/p>

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <set>
#include <vector>
using namespace std;
typedef long long ll;
typedef long double ld;
namespace 
struct Line 
mutable ll k, m, p;
bool f; // 瀛樺湪鏂滅巼鍚?/span>
Line() 
Line(ll _k, ll _m, ll _p, bool _f) : k(_k), m(_m), p(_p), f(_f)  
bool friend operator < (const Line &a, const Line &b) 
return (a.f && b.f) ? a.k < b.k : a.p < b.p;

;
struct LineContainer : multiset<Line> 
//    LineContainer() 
const ll inf = LLONG_MAX;
ll div(ll a, ll b)  //姹備氦鐐?/span>
return a / b - (a ^ b < 0 && a % b); 

ld div(ld a, ld b) 
return a / b;

bool Intersect(iterator x, iterator y) 
if(y == end()) 
x -> p = inf;
return false;

if(x -> k == y -> k) x -> p = x -> m > y -> m ? inf : -inf;
else x -> p = div(y -> m - x -> m, x -> k - y -> k);
return x -> p >= y -> p;

void add(ll k, ll m) 
multiset<Line> :: iterator z = insert(Line(k, m, 0, 1)), y = z++, x = y;
while(Intersect(y, z)) z = erase(z);
if(x != begin() && Intersect(--x, y)) Intersect(x, y = erase(y));
while((y = x) != begin() && (--x) -> p >= y -> p) Intersect(x, erase(y));

ll query(ll x) 
//    assert(!empty());
multiset<Line> :: iterator L = lower_bound(Line(0, 0, x, 0));
return L -> k * x + L -> m;

;

const int maxn = 5e5 + 5;
int n;
ll tot, ans;
ll a[maxn], sum[maxn];
LineContainer H;
int main() 
scanf("%d", &n);
for(int i = 1; i <= n; ++i) 
scanf("%lld", &a[i]);
tot += a[i] * i;
sum[i] = sum[i - 1] + a[i];
    
ans = tot;
for(int i = 1; i <= n; ++i) 
if(!H.empty()) ans = max(ans, H.query(a[i]) + sum[i - 1] + tot - a[i] * i);
H.add(1.0 * i, 1.0 * -sum[i - 1]);

H.clear();
for(int i = n; i; --i) 
if(!H.empty()) ans = max(ans, H.query(a[i]) + sum[i] + tot - a[i] * i);
H.add(1.0 * i, 1.0 * -sum[i]);
    
printf("%lld\n", ans);
return 0;

/*
2 3 4 5 6
5 2 3 4 6
j = 2 i = 5 
鍓嶅悗鍋氫袱閬?
鍓? 鍚戝墠绉籭绉诲埌j鍓嶉潰
ans = max(tot + a[i] * j - a[i] * i + (sum[i - 1] - sum[j - 1]))
= max(tot + a[i] * j - sum[i - 1] - a[i] * i + sum[j - 1]) j <= i
y = k * x + b
y 鏈€澶?
ans - tot - sum[i - 1] + a[i] * i = a[i] * j - sum[j - 1]
x = a[i]
y = ans - tot - sum[i - 1] + a[i] * i
k = j
b = -sum[j - 1]
缁存姢涓嬪嚫鍖?
鍚?
2 3 4 5 6
3 4 5 2 6
ans = max(tot + a[i] * j - a[i] * i + (sum[i] - sum[j]))
*/

View Code