HDU 4816 Bathysphere （2013长春现场赛D题）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4816

2013长春区域赛的D题。

很简单的几何题，就是给了一条折线。 然后一个矩形窗去截取一部分，求最大面积。

现场跪在这题，最后时刻TLE到死，用的每一小段去三分，时间复杂度是O(n log n) , 感觉数据也不至于超时。

卧槽！！！！代码拷回来，今天在HDU一交，一模一样的代码AC了，加输入外挂6s多，不加也8s多，都可AC，呵呵·····（估计HDU时限放宽了！！！）

现场赛卡三分太SXBK了，我艹！！！！ 好好的一个现场赛绝杀错过了。

以下是现场赛源码，在HDU顺利AC！ 现场TLE到死！

其实O(n)也可以搞，因为是三分的是一个二次函数，求对称轴就可以了。现场没有想到这个优化，等下简单修改成O(n)代码。

 #include <stdio.h>
 #include <string.h>
 #include <iostream>
 #include <algorithm>
 #include <queue>
 #include <map>
 #include <set>
 #include <vector>
 #include <string>
 #include <math.h>
 using namespace std;
 const int MAXN = ;
 int x[MAXN],y[MAXN];
 int d;
 int n;
 int L;
 int nowx ;
 int nextx;
 int r1,l2;
 const double eps = 1e-;
 double solve()
 {
     double left = nowx,right = nextx;
     double ret1,ret2;
     for(int cc = ;cc <= ;cc++)
     {
         double mid = (left + right)/;
         double midmid = (mid + right)/;
         double h1 = y[r1] + (double)(y[r1-] - y[r1]) * (mid  - x[r1])/(x[r1-] - x[r1]);
         double h2 = y[l2] + (double)(y[l2+] - y[l2])*(mid + *d - x[l2])/(x[l2 + ] - x[l2]);
         ret1 = (double)(x[r1] - mid)*(h1 + y[r1])/ + (double)(mid + *d - x[l2])*(h2 + y[l2])/;

          h1 = y[r1] + (double)(y[r1-] - y[r1]) * (midmid  - x[r1])/(x[r1-] - x[r1]);
          h2 = y[l2] + (double)(y[l2+] - y[l2])*(midmid + *d - x[l2])/(x[l2 + ] - x[l2]);
         ret2 = (double)(x[r1] - midmid)*(h1 + y[r1])/ + (double)(midmid + *d - x[l2])*(h2 + y[l2])/;
         if(ret1 < ret2)
             left = mid+eps;
         else right = midmid-eps;
     }
     return ret1;
 }

 int input()
 {
     char ch;
     ch = getchar();
     while(ch < '' || ch >'')
     {
         ch = getchar();
     }
     int ret = ;
     while(ch >= '' && ch <= '')
     {
         ret *= ;
         ret += ch -'';
         ch = getchar();
     }
     return ret;
 }

 int main()
 {
     //freopen("D.txt","r",stdin);
     //freopen("out.txt","w",stdout);
     int T;
     scanf("%d",&T);
     while(T--)
     {
         scanf("%d%d",&n,&L);
         for(int i = ;i <= n;i++)
         {
             //x[i] = input();
             //y[i] = input();
             scanf("%d%d",&x[i],&y[i]);
         }
             //scanf("%d%d",&x[i],&y[i]);
         scanf("%d",&d);
         double ans = ;
          r1 = ;
          l2 = ;
         double tmp = ;
         while(l2 < n && x[l2+] < *d)l2++;
         for(int i = r1;i < l2;i++)
         {
             tmp += (double)(x[i+] - x[i])*(y[i] + y[i+])/;
         }
         if(l2  == )
         {
             tmp -= (double)(x[] - x[])*(y[] + y[])/;
         }
         x[n+] = x[n];
         y[n+] = y[n];
         nowx = ;
         //printf("%d %d\n",r1,l2);
         while(l2 < n && r1 <= n)
         {
             int p1 = x[r1];
             int p2 = x[l2 + ] - *d;
             if(p1 < p2)
                 nextx = p1;
             else nextx = p2;
             nextx = min(L- *d,nextx);
             //printf("%d %d\n",nowx,nextx);
             ans = max(ans,tmp + solve());
             if(p1 < p2)
             {
                 nowx = p1;
                 if(r1 < n)tmp -= (double)(x[r1+] - x[r1])*(y[r1+] + y[r1] )/;
                 r1++;
             }
             else
             {
                 nowx = p2;
                 tmp += (double)(x[l2+] - x[l2])*(y[l2+] + y[l2])/;
                 l2++;
             }
         }
         printf("%.3lf\n",ans//d);
     }
     return ;
 }

改成了O(n)的解法。

因为三分的那个函数是二次函数。

找最大值，只要找两个端点和对称轴处就足够了！

还是太弱，现场没有想到这个优化，改起来很容易的。

 #include <stdio.h>
 #include <string.h>
 #include <iostream>
 #include <algorithm>
 #include <queue>
 #include <map>
 #include <set>
 #include <vector>
 #include <string>
 #include <math.h>
 using namespace std;
 const int MAXN = ;
 int x[MAXN],y[MAXN];
 int d;
 int n;
 int L;
 int nowx ;
 int nextx;
 int r1,l2;
 const double eps = 1e-;
 double solve()
 {
     double left = nowx,right = nextx;
     //求出二次函数的a,b,c系数
     double tmp11 = x[r1];
     double tmp12 = -;
     double tmp13 = (double)y[r1] - (double)x[r1]*(y[r1-] - y[r1])/(x[r1-] - x[r1])/;
     double tmp14 = (double)(y[r1-] - y[r1])/(x[r1-] - x[r1])/;
     double tmp21 = *d - x[l2];
     double tmp22 = ;
     double tmp23 = y[l2] + (double)(*d - x[l2])*(y[l2+] - y[l2])/(x[l2+] - x[l2])/;
     double tmp24 = (double)(y[l2+] - y[l2])/(x[l2+] - x[l2])/;
     //函数Y = (tmp11 + tmp12 * x)*(tmp13 + tmp14 * x) + (tmp21 + tmp22 * x)*(tmp23 + tmp24*x)
     double a = tmp12 * tmp14 + tmp22 * tmp24;
     double b = tmp11 * tmp14 + tmp12 * tmp13 + tmp21 * tmp24 + tmp22 * tmp23;
     double c = tmp11 * tmp13 + tmp21 * tmp23;

     double x0 = -b /(*a);//对称轴
     double ret = max(a*left*left + b*left + c,a*right *right + b * right + c);
     if(x0 >= left && x0 <= right)
         ret = max(ret,a * x0 * x0 + b*x0 + c);
     return ret;

     /*
     double ret1,ret2;
     for(int cc = 0;cc <= 30;cc++)
     {
         double mid = (left + right)/2;
         double midmid = (mid + right)/2;
         double h1 = y[r1] + (double)(y[r1-1] - y[r1]) * (mid  - x[r1])/(x[r1-1] - x[r1]);
         double h2 = y[l2] + (double)(y[l2+1] - y[l2])*(mid + 2*d - x[l2])/(x[l2 + 1] - x[l2]);
         ret1 = (double)(x[r1] - mid)*(h1 + y[r1])/2 + (double)(mid + 2*d - x[l2])*(h2 + y[l2])/2;

          h1 = y[r1] + (double)(y[r1-1] - y[r1]) * (midmid  - x[r1])/(x[r1-1] - x[r1]);
          h2 = y[l2] + (double)(y[l2+1] - y[l2])*(midmid + 2*d - x[l2])/(x[l2 + 1] - x[l2]);
         ret2 = (double)(x[r1] - midmid)*(h1 + y[r1])/2 + (double)(midmid + 2*d - x[l2])*(h2 + y[l2])/2;
         if(ret1 < ret2)
             left = mid+eps;
         else right = midmid-eps;
     }
     return ret1;
     */
 }

 int input()
 {
     char ch;
     ch = getchar();
     while(ch < '' || ch >'')
     {
         ch = getchar();
     }
     int ret = ;
     while(ch >= '' && ch <= '')
     {
         ret *= ;
         ret += ch -'';
         ch = getchar();
     }
     return ret;
 }

 int main()
 {
     //freopen("D.txt","r",stdin);
     //freopen("out.txt","w",stdout);
     int T;
     scanf("%d",&T);
     while(T--)
     {
         scanf("%d%d",&n,&L);
         for(int i = ;i <= n;i++)
         {
             //x[i] = input();
             //y[i] = input();
             scanf("%d%d",&x[i],&y[i]);
         }
             //scanf("%d%d",&x[i],&y[i]);
         scanf("%d",&d);
         double ans = ;
          r1 = ;
          l2 = ;
         double tmp = ;
         while(l2 < n && x[l2+] < *d)l2++;
         for(int i = r1;i < l2;i++)
         {
             tmp += (double)(x[i+] - x[i])*(y[i] + y[i+])/;
         }
         if(l2  == )
         {
             tmp -= (double)(x[] - x[])*(y[] + y[])/;
         }
         x[n+] = x[n];
         y[n+] = y[n];
         nowx = ;
         //printf("%d %d\n",r1,l2);
         while(l2 < n && r1 <= n)
         {
             int p1 = x[r1];
             int p2 = x[l2 + ] - *d;
             if(p1 < p2)
                 nextx = p1;
             else nextx = p2;
             nextx = min(L- *d,nextx);
             //printf("%d %d\n",nowx,nextx);
             ans = max(ans,tmp + solve());
             if(p1 < p2)
             {
                 nowx = p1;
                 if(r1 < n)tmp -= (double)(x[r1+] - x[r1])*(y[r1+] + y[r1] )/;
                 r1++;
             }
             else
             {
                 nowx = p2;
                 tmp += (double)(x[l2+] - x[l2])*(y[l2+] + y[l2])/;
                 l2++;
             }
         }
         printf("%.3lf\n",ans//d);
     }
     return ;
 }