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 k ₁, k ₂, … k _n, a ₁, a ₂, …, a _n and b ₁, b ₂, …, 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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liuyiding

 

 

数值方法计算积分

 

  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\n",ans);

124     }

125     return 0;

126 }