HDU 4498 Function Curve （自适应simpson）

Function Curve

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 203 Accepted Submission(s): 67

Problem Description

Given sequences of k₁, k₂, … k_n, a₁, a₂, …, a_n and b₁, b₂, …, b_n. Consider following function:

Then we draw F(x) on a xy-plane, the value of x is in the range of [0,100]. Of course, we can get a curve from that plane.
Can you calculate the length of this curve？

Input

The first line of the input contains one integer T (1<=T<=15), representing the number of test cases.
Then T blocks follow, which describe different test cases.
The first line of a block contains an integer n ( 1 <= n <= 50 ).
Then followed by n lines, each line contains three integers k_i, a_i, b_i ( 0<=a_i, b_i<100, 0<k_i<100 ) .

Output

For each test case, output a real number L which is rounded to 2 digits after the decimal point, means the length of the curve.

Sample Input

2 3 1 2 3 4 5 6 7 8 9 1 4 5 6

Sample Output

215.56 278.91

Hint

All test cases are generated randomly.

Source

2013 ACM-ICPC吉林通化全国邀请赛——题目重现

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  1 /* ***********************************************
  2 Author        :kuangbin
  3 Created Time  :2013-10-9 12:04:07
  4 File Name     :E:2013ACM专题学习数学积分HDU4498.cpp
  5 ************************************************ */
  6 
  7 #include <stdio.h>
  8 #include <string.h>
  9 #include <iostream>
 10 #include <algorithm>
 11 #include <vector>
 12 #include <queue>
 13 #include <set>
 14 #include <map>
 15 #include <string>
 16 #include <math.h>
 17 #include <stdlib.h>
 18 #include <time.h>
 19 using namespace std;
 20 
 21 vector<double>x;
 22 
 23 void add(int a1,int b1,int c1)//计算a1*x^2 + b1*x + c = 0的解
 24 {
 25     if(a1 == 0 && b1 == 0)
 26     {
 27         return;
 28     }
 29     if(a1 == 0)
 30     {
 31         double t = -c1*1.0/b1;
 32         if(t >= 0 && t <= 100)
 33             x.push_back(t);
 34         return;
 35     }
 36     long long deta = b1*b1 - 4LL*a1*c1;
 37     if(deta < 0)return;
 38     if(deta == 0)
 39     {
 40         double t = (-1.0 * b1)/(2.0 * a1);
 41         if(t >= 0 && t <= 100)
 42             x.push_back(t);
 43     }
 44     else
 45     {
 46         double t1 = (-1.0 * b1 + sqrt(1.0*deta))/(2.0*a1);
 47         double t2 = (-1.0 * b1 - sqrt(1.0*deta))/(2.0*a1);
 48         if(t1 >= 0 && t1 <= 100)
 49             x.push_back(t1);
 50         if(t2 >= 0 && t2 <= 100)
 51             x.push_back(t2);
 52     }
 53 }
 54 int A[100],B[100],C[100];
 55 int best;
 56 double F(double x1)
 57 {
 58     return sqrt(1.0 + (x1*2*A[best] + 1.0 * B[best])*(x1*2*A[best] + 1.0 * B[best]));
 59 }
 60 double simpson(double a,double b)
 61 {
 62     double c = a + (b-a)/2;
 63     return (F(a) + 4*F(c) + F(b))*(b-a)/6;
 64 }
 65 double asr(double a,double b,double eps,double A)
 66 {
 67     double c = a + (b-a)/2;
 68     double L = simpson(a,c);
 69     double R = simpson(c,b);
 70     if(fabs(L+R-A) <= 15*eps)return L+R+(L+R-A)/15;
 71     return asr(a,c,eps/2,L) + asr(c,b,eps/2,R);
 72 }
 73 double asr(double a,double b,double eps)
 74 {
 75     return asr(a,b,eps,simpson(a,b));
 76 }
 77 
 78 int main()
 79 {
 80     //freopen("in.txt","r",stdin);
 81     //freopen("out.txt","w",stdout);
 82     int T;
 83     int k,a,b;
 84     scanf("%d",&T);
 85     while(T--)
 86     {
 87         int n;
 88         scanf("%d",&n);
 89         A[0] = 0;
 90         B[0] = 0;
 91         C[0] = 100;
 92         for(int i = 1;i <= n;i++)
 93         {
 94             scanf("%d%d%d",&k,&a,&b);
 95             A[i] = k;
 96             B[i] = -2*a*k;
 97             C[i] = k*a*a + b;
 98         }
 99         x.clear();
100         for(int i = 0;i <= n;i++)
101             for(int j = i+1;j <= n;j++)
102                 add(A[i]-A[j],B[i] - B[j],C[i] - C[j]);
103         double ans = 0;
104         x.push_back(0);
105         x.push_back(100);
106         sort(x.begin(),x.end());
107         int sz = x.size();
108         for(int i = 0;i < sz-1;i++)
109         {
110             double x1 = x[i];
111             double x2 = x[i+1];
112             if(fabs(x2-x1) < 1e-8)continue;
113             double mid = (x1 + x2)/2;
114             best = 0;
115             for(int j = 1;j <= n;j++)
116             {
117                 double tmp1 = mid*mid*A[best] + mid*B[best] + C[best];
118                 double tmp2 = mid*mid*A[j] + mid*B[j] + C[j];
119                 if(tmp2 < tmp1)best = j;
120             }
121             ans += asr(x1,x2,1e-8);
122         }
123         printf("%.2lf
",ans);
124     }
125     return 0;
126 }

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原文地址：https://www.cnblogs.com/kuangbin/p/3358984.html