很容易得出轉移方程，f[i]=max/(rate[i]*a[i]+b[i])，其中f[i]表示在第i天所獲得的的最多的b卷
然後，我們發現會超時，我們稍微化簡一下，由f[i] = (rate[j]*f[j]*a[i]+f[j]*b[i])/(rate[i]*a[i]+b[i])得
f[i]*(rate[i]*a[i]+b[i])/a[i] = (-b[i]/a[i])*f[j]+f[j]*rate[j]
相當於我們需要維護乙個資料結構，滿足
1.插入乙個點(f[i],f[i]*rate[i])
2.給定負斜率-b[i]/a[i],求最大截距
嗯，平衡樹上吧
#include #include #include #include using namespace std;
const int maxn = 100001;
const double inf = 10000001;
int n;
double s;
double f[maxn];
double a[maxn],b[maxn],rate[maxn];
double ans;
bool f;
struct point
point() {}
point(double _x,double _y,double _k):x(_x),y(_y),k(_k){}
};typedef point vector;
vector operator + (vector a,vector b);}
vector operator - (point a,point b);}
inline double cross(vector a,vector b)
set s;
set :: iterator it1,it2;
inline double slope(point a,point b)
inline double cut(double k,point p)
bool judge(point p)
void insert(point p)
it1 = it2 = s.lower_bound(p);
it2++;
while (it1 != s.end() && it2 != s.end())
it1 = it2 = s.lower_bound(p);
if (it1 != s.begin())
if (it2 != s.end())
s.insert(p);
point temp = *(--s.end());
temp.k = inf;
s.erase(--s.end()), s.insert(temp);
}int main()
); for (int i=2;i<=n;i++)
); f[i] = max(ans,cut(-b[i]/a[i],temp)*a[i])/(rate[i]*a[i]+b[i]);
ans = max(ans,rate[i]*f[i]*a[i]+f[i]*b[i]);
temp = *--s.lower_bound((point));
f[i] = max(ans,cut(-b[i]/a[i],temp)*a[i])/(rate[i]*a[i]+b[i]);
ans = max(ans,rate[i]*f[i]*a[i]+f[i]*b[i]);
insert((point));
} printf("%.3lf\n",ans);
return 0;
}
鑑於那個時候還沒有set，splay寫起來就是要死要死的，那麼我們還可以使用cdq分治，說起來簡單，還不如直接寫動態凸包
#include #include #include #include #include using namespace std;
const int maxn = 100001;
const double inf = 10000001;
int n;
double s;
double f[maxn];
double a[maxn],b[maxn],rate[maxn];
double ans;
bool f;
struct point
};typedef point vector;
vector operator + (vector a,vector b);}
vector operator - (point a,point b);}
inline double cross(vector a,vector b)
int convex_size;
point p[maxn];
point convex[maxn];
inline double slope(point a,point b)
inline double cut(double k,point p)
int find(double k)
return l;
}void solve(int l,int r)
; return;
} int mid = (l+r)/2;
solve(l,mid);
convex_size = 0;
for (int i=l;i<=mid;i++)
convex[++convex_size] = (point);
for (int i=mid+1;i<=r;i++)
solve(mid+1,r);
inplace_merge(p+l,p+mid+1,p+r+1);
}int main()
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