题目链接
题解
首先，状态转移方程
\\(f_i = max(f_j+A(S_i-S_j)^2+B(S_i-S_j)+C)\\)
在这里总结一下推斜率优化的两种方法吧
直接推呀:
设\\(j<k\\)且\\(j\\)比\\(k\\)优。
\\[f_j+A(S_i-S_j)^2+B(S_i-S_j)+C>f_k+A(S_i-S_k)^2+B(S_i-S_k)+C
\\]
\\[f_j-f_k+A(2S_i-S_j-S_k)*(S_k-S_j)+B(S_k-S_j)>0
\\]
\\[f_j+S_j^2-2AS_iS_j-BS_j >f_k+S_k^2-2AS_iS_k-BS_k
\\]
设\\(G_j=f_j+AS_j^2-BS_j\\),\\(H_j=-2AS_j\\)
得到
\\(\\frac{G_j-G_k}{H_j-H_k}<-S_i\\)
找直线解析式
代码
#include<cstdio>
#include<cstring>
#include<algorithm>
#define LL long long
inline LL read() {
LL x = 0,f = 1;
char c = getchar();
while(c < \'0\' || c > \'9\'){ if(c == \'-\')f = -1; c = getchar();}
while(c <= \'9\' && c >= \'0\')x = x * 10 + c - \'0\',c = getchar();
return x * f;
}
const int maxn = 1000007;
LL n,a,b,c;
LL sum[maxn];
LL dp[maxn];
LL Y (int i) { return dp[i] + a * sum[i] * sum[i] - b * sum[i]; }
LL X (int i) { return sum[i]; }
double slop(int i,int j) { return 1.0 * (Y(i) - Y(j)) / (X(i) - X(j)); }
int q[maxn];
int main() {
n = read(),a = read(),b = read(),c = read();
for(int i = 1;i <= n;++ i) sum[i] = read() + sum[i - 1];
for(int l = 0,r = 0,i = 1;i <= n;++ i) {
while(l < r && slop(q[l],q[l + 1]) > 2 * a * sum[i])l ++;
//dp[i] = dp[q[l]] + a * (sum[i] - sum[q[l]]) * (sum[i] - sum[q[l]]) + b * (sum[i] - sum[q[l]] + c);
dp[i] = -(2 * a * sum[i] * X(q[l]) - Y(q[l]) - a * sum[i] * sum[i] - b * sum[i] - c);
while(l < r && slop(q[r - 1],q[r]) <= slop(q[r],i)) r --;
q[++ r] = i;
}
printf("%lld\\n",dp[n]);
return 0;
}