【wqs二分】HHHOJ#15. 赤

这个wqs二分并不熟练……

题目描述

#15. 赤

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题目分析

两维都用wqs二分，其他没有什么特殊之处。

重点在于，wqs二分还原最优解的时候，增量是强制给的k。

 #include<bits/stdc++.h>
 const int maxn = ;
 const double eps = 1e-;

 int n,a,b,usa,usb;
 double bet,ans,coa,cob,p[maxn],q[maxn];

 bool match()
 {
     double k1=coa,k2=cob;
     usa = usb = , ans = ;
     for (int i=, opt=; i<=n; i++)
     {
         double v = ;opt=;
         if (p[i]-k1 > v) opt = , v = p[i]-k1;
         if (q[i]-k2 > v) opt = , v = q[i]-k2;
         if (p[i]+q[i]-p[i]*q[i]-k1-k2 > v)
             opt = , v = p[i]+q[i]-p[i]*q[i]-k1-k2;
         if (opt==||opt==) usa++;
         if (opt==||opt==) usb++;
         ans += v;
     }
     return usb <= b;
 }
 bool check()
 {
     double l = , r = 1.0, bet;
     for (cob=(l+r)/2.0; r-l>eps; cob=(l+r)/2.0)
         if (match()) bet = r = cob;
         else l = cob;
     cob = bet;
     match();
     return usa <= a;
 }
 int main()
 {
     while (scanf("%d%d%d",&n,&a,&b)!=EOF)
     {
         for (int i=; i<=n; i++) scanf("%lf",&p[i]);
         for (int i=; i<=n; i++) scanf("%lf",&q[i]);
         double l = , r = 1.0;
         for (coa=(l+r)/2.0; r-l>eps; coa=(l+r)/2.0)
             if (check()) bet = coa, r = coa;
             else l = coa;
         coa = bet;
         check();
         ans += 1.0*a*coa+1.0*b*cob;　　//这里补偿答案用a,b
         printf("%.5lf\n",ans);
     }
     return ;
 }

END