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

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)

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#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[N]);
return 0;
}

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## 1911: [Apio2010]特别行动队

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

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## Source

posted @ 2015-09-15 21:00  JSZX11556  阅读(310)  评论(0编辑  收藏  举报