您的位置:首页 > 其它

BZOJ 1911: [Apio2010]特别行动队( dp + 斜率优化 )

2015-09-15 21:00 323 查看


sum为战斗力的前缀和

dp(x) = max( dp(p)+A*(sumx-sump)2+B*(sumx-sump)+C )(0≤p<x)

然后斜率优化...懒得写下去了...

--------------------------------------------------------------------------------

#include<bits/stdc++.h> using namespace std; typedef long long ll; const int maxn = 1000009; int N, A, B, C, Q[maxn], qh, qt;ll w[maxn], dp[maxn]; inline ll f(int x) { return dp[x] + ll(A) * w[x] * w[x];} inline ll g(int x, int p) { return dp[p] + ll(A) * (w[x] - w[p]) * (w[x] - w[p]) + ll(B) * (w[x] - w[p]) + C;} int main() { scanf("%d%d%d%d", &N, &A, &B, &C); w[0] = 0; for(int i = 1; i <= N; i++) { scanf("%d", w + i); w[i] += w[i - 1]; } qh = qt = 0; Q[qt++] = 0; dp[0] = 0; for(int i = 1; i <= N; i++) { ll t = 2LL * A * w[i] + B; while(qt - qh >= 2 && f(Q[qh + 1]) - f(Q[qh]) > t * (w[Q[qh + 1]] - w[Q[qh]])) qh++; dp[i] = g(i, Q[qh]); while(qt - qh >= 2 && (ll) (f(i) - f(Q[qt - 1])) * (w[Q[qt - 1]] - w[Q[qt - 2]]) > (ll) (f(Q[qt - 1]) - f(Q[qt - 2])) * (w[i] - w[Q[qt - 1]])) qt--; Q[qt++] = i; } printf("%lld\n", dp
); return 0;}--------------------------------------------------------------------------------

1911: [Apio2010]特别行动队

Time Limit: 4 Sec Memory Limit: 64 MB
Submit: 2998 Solved: 1354
[Submit][Status][Discuss]

Description


Input


Output


Sample Input

4
-1 10 -20
2 2 3 4

Sample Output

9

HINT



Source

内容来自用户分享和网络整理,不保证内容的准确性,如有侵权内容,可联系管理员处理 点击这里给我发消息
标签: