[APIO2010]特別行動隊

題面很直白，就不放了。

太套路了，做起來沒點感覺了。

\(dp(i)=dp(j)+a*(s(i)-s(j))^{2}+b*(s(i)-s(j))+c\)

直接推出一個斜率優化的式子上單調隊列就好了

時間/空間復雜度：\(O(n)\)

#include<cstdio>
#define sid 1000500
#define ri register int
#define ll long long
#define dd double
using namespace std;

#define getchar() *S ++
char RR[30000005], *S = RR;
inline 
 
int read(){
    int p = 0, w = 1;
    char c = getchar();
    while(c > ‘9‘ || c < ‘0‘) {
        if(c == ‘-‘) w = -1;
        c =  getchar();
    }
    while(c >= ‘0‘ && c <= ‘9‘) {
        p = p * 10 + c - ‘0‘;
        c = getchar();
    }
    return p * w;
}

ll dp[sid], sum[sid];
 
int q[sid], n, a, b, c;

#define x(g) (sum[(g)])
#define y(g) (dp[(g)] + a * sum[(g)] * sum[(g)] - b * sum[(g)])

inline dd s(int i,int j){
    return (dd)(y(i) - y(j)) / (dd)(x(i) - x(j));
}

int main(){
    fread(RR, 1, sizeof(RR), stdin);
    n = read(); a = read();
    b = read(); c = read();
    
 
for(ri i = 1; i <= n; i ++) sum[i] = sum[i - 1] + read();
    ri fr = 1, to = 1;
    for(ri i = 1; i <= n; i ++){
        while(fr + 1 <= to && s(q[fr],q[fr + 1]) > 2 * a * sum[i]) fr ++;
        int p = q[fr];
        ll pp = sum[i] - sum[p];
        dp[i] = dp[p] + a * pp * pp + b * pp + c;
        while(fr + 1 <= to && s(q[to], q[to - 1]) <= s(i, q[to - 1])) to --;
        q[++ to]=i;
    }
    printf("%lld\n", dp[n]);
    return 0;
}

[APIO2010]特別行動隊 --- 斜率優化DP