BZOJ 3156: 防御准备( dp + 斜率优化 )

dp(i)表示处理完[i,n]且i是放守卫塔的最小费用.

dp(i) = min{dp(j) + (j-i)(j-i-1)/2}+cost_i(i<j≤N)

然后斜率优化

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#include<cstdio>

#include<cstring>

#include<algorithm>

using namespace std;

typedef long long ll;

#define f(x) (dp[x] * 2 + ll(x) * x)

const int maxn = 1000009;

ll dp[maxn];

int cost[maxn], N;

int q[maxn], qh = 0, qt = -1;

double K(int x, int y) {

return (double) (f(x) - f(y)) / (x - y);

}

int main() {

scanf("%d", &N);

for(int i = 1; i <= N; i++)

scanf("%d", cost + i);

cost[0] = 0;

dp[N] = cost[N];

q[++qt] = N;

for(int i = N; i--; ) {

while(qt - qh > 0 && K(q[qh], q[qh + 1]) > i * 2 + 1) qh++;

dp[i] = dp[q[qh]] + 1LL * (q[qh] - i) * (q[qh] - i - 1) / 2 + cost[i];

while(qt - qh > 0 && K(q[qt], i) > K(q[qt - 1], q[qt])) qt--;

q[++qt] = i;

}

printf("%lld ", dp[0]);

return 0;

}

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3156: 防御准备

Time Limit: 10 Sec Memory Limit: 512 MB
Submit: 788 Solved: 375
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Description

Input

第一行为一个整数N表示战线的总长度。

第二行N个整数，第i个整数表示在位置i放置守卫塔的花费Ai。

Output

共一个整数，表示最小的战线花费值。

Sample Input

10
2 3 1 5 4 5 6 3 1 2

Sample Output

HINT

1<=N<=10^6,1<=Ai<=10^9

Source

Katharon+#1