题目传送门
1911: [Apio2010]特别行动队
Time Limit: 4 Sec Memory Limit: 64 MBSubmit: 5349 Solved: 2651
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Description
Input
Output
Sample Input
4
-1 10 -20
2 2 3 4
-1 10 -20
2 2 3 4
Sample Output
9
HINT
Source
斜率优化dp
不过需要注意的是这道题斜率是负的，并且维护的是个上凸包。
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #include<cmath> 6 #define LL long long 7 using namespace std ; 8 const int N = 1e6 + 10 ; 9 10 inline LL read() { 11 LL k = 0, f = 1 ; char c = getchar() ; 12 for( ; !isdigit(c) ; c = getchar()) 13 if(c == '-') f = -1 ; 14 for( ; isdigit(c) ; c = getchar()) 15 k = k*10 + c-'0' ; 16 return k*f ; 17 } 18 int n ; LL a, b, c ; LL pp[N] ; 19 int q[N] ; LL f[N] ; 20 21 inline LL sqr(LL x) { return x*x ; } 22 inline double Y(int i) { return f[i]+a*sqr(pp[i])-b*pp[i] ; } 23 inline double X(int i) { return pp[i] ; } 24 inline double Rate(int i,int j) { return (Y(j)-Y(i))/(X(j)-X(i)) ; } 25 26 int main() { 27 n = read() ; a = read(), b = read(), c = read() ; 28 for(int i=1;i<=n;i++) pp[i] = pp[i-1]+read() ; 29 int head = 1, tail = 1 ; q[1] = 0 ; 30 for(int i=1;i<=n;i++) { 31 while(head < tail && Rate(q[head],q[head+1]) > (double)2*a*pp[i]) head++ ; 32 int j = q[head] ; f[i] = f[j]+a*sqr(pp[i]-pp[j])+b*(pp[i]-pp[j])+c ; 33 while(head < tail && Rate(q[tail-1],q[tail]) < Rate(q[tail],i)) tail-- ; 34 q[++tail] = i ; 35 } 36 printf("%lld",f[n]) ; 37 return 0 ; 38 }