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*(sum_x-sum_p)²+B*(sum_x-sum_p)+C )(0≤p<x)

然后斜率优化...懒得写下去了...

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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
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Description

Input

Output

Sample Input

4
-1 10 -20
2 2 3 4

巴特西

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

1911: [Apio2010]特别行动队

Description

Input

Output

Sample Input

Sample Output

HINT

Source

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