[APIO2010]特別行動隊
阿新 • • 發佈:2018-12-17
pro memset ace span clas git reg ack efi ,那麽
\(y = dp[j] - b * s[j] + a * s[j] ^ 2\),
\(k = 2a * s[i]\),
\(x = s[j]\),
\(b = dp[i] - a * s[i] ^ 2 - b * s[i] - c\)。
方便的是\(x\)是單調遞增的,\(k\)是單調遞減的,於是用最樸素的單調隊列維護上凸包即可。
斜率優化寫的還是不太熟練
嘟嘟嘟
這道題dp式特別好想:
\[dp[i] = max_{j = 0} ^ {i - 1} (dp[j] + f(s[i] - s[j]))\]
其中\(f(x) = ax^ 2 + bx + c\),\(s[i] = \sum_{j = 1} ^ {i} x[j]\)。
但是\(O(n ^ 2)\)過不了,需要斜率優化。
勇敢的把項拆開得到
\[dp[i] = max(dp[j] - b * s[j] + a * s[j] ^ 2 - 2a * s[i] * s[j]) + a * s[i] ^ 2 + b * s[i] + c\]
然後就是套路:把這個式子看成\(b = y - k * x\)
\(y = dp[j] - b * s[j] + a * s[j] ^ 2\),
\(k = 2a * s[i]\),
\(x = s[j]\),
\(b = dp[i] - a * s[i] ^ 2 - b * s[i] - c\)。
方便的是\(x\)是單調遞增的,\(k\)是單調遞減的,於是用最樸素的單調隊列維護上凸包即可。
斜率優化寫的還是不太熟練
#include<cstdio> #include<iostream> #include<cmath> #include<algorithm> #include<cstring> #include<cstdlib> #include<cctype> #include<vector> #include<stack> #include<queue> using namespace std; #define enter puts("") #define space putchar(‘ ‘) #define Mem(a, x) memset(a, x, sizeof(a)) #define rg register typedef 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; }
[APIO2010]特別行動隊