这道题dp式特别好想：
\[dp[i] = max_{j = 0} ^ {i - 1} (dp[j] + f(s[i] - s[j]))\] 其中
\(f(x) = ax^ 2 + bx + c\)，
\(s[i] = \sum_{j = 1} ^ {i} x[j]\)。
但是
\(O(n ^ 2)\)过不了，需要斜率优化。
勇敢的把项拆开得到
\[dp[i] = max(dp[j] - b * s[j] + a * s[j] ^ 2 - 2a * s[i] * s[j]) + a * s[i] ^ 2 + b * s[i] + c\] 然后就是套路：把这个式子看成
\(b = y - k * x\)，那么
\(y = dp[j] - b * s[j] + a * s[j] ^ 2\),
\(k = 2a * s[i]\),
\(x = s[j]\),
\(b = dp[i] - a * s[i] ^ 2 - b * s[i] - c\)。
方便的是
\(x\)是单调递增的，
\(k\)是单调递减的，于是用最朴素的单调队列维护上凸包即可。
斜率优化写的还是不太熟练
#include #include #include #include #include #include #include #include #include #include using namespace std;#define enter puts("") #define space putchar(' ')#define Mem(a, x) memset(a, x, sizeof(a))#define rg registertypedef long long ll;typedef double db;const int INF = 0x3f3f3f3f;const db eps = 1e-8;const int maxn = 1e6 + 5;inline ll read(){ ll ans = 0; char ch = getchar(), last = ' '; while(!isdigit(ch)) {last = ch; ch = getchar();} while(isdigit(ch)) {ans = (ans << 1) + (ans << 3) + ch - '0'; ch = getchar();} if(last == '-') ans = -ans; return ans;}inline void write(ll x){ if(x < 0) x = -x, putchar('-'); if(x >= 10) write(x / 10); putchar(x % 10 + '0');}int n, a, b, c;ll sum[maxn], dp[maxn];int q[maxn], l = 0, r = 0;#define x(i) sum[i]#define k(i) (2 * a * sum[i])#define y(i) (dp[i] - b * sum[i] + a * sum[i] * sum[i])db slope(int i, int j){ return 1.0 * (y(i) - y(j)) / (x(i) - x(j));}int main(){ n = read(); a = read(), b = read(), c = read(); for(int i = 1, x; i <= n; ++i) x = read(), sum[i] = sum[i - 1] + x; for(int i = 1; i <= n; ++i) { while(l < r && slope(q[l], q[l + 1]) > k(i)) l++; dp[i] = y(q[l]) - k(i) * x(q[l]) + a * sum[i] * sum[i] + b * sum[i] + c; while(l < r && slope(q[r - 1], q[r]) <= slope(q[r], i)) r--; q[++r] = i; } write(dp[n]), enter; return 0;}