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HDU 4498 Function Curve （分段，算曲线积分）

HDU Function 曲线 积分 分段
2023-09-27 14:26:06 时间

Function Curve

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 31    Accepted Submission(s): 10


Problem Description
Given sequences of k1, k2, … kn, a1, a2, …, an and b1, b2, …, bn. 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 ki, ai, bi ( 0<=ai, bi<100, 0<ki<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
 

 

Recommend
liuyiding
 

 

 

 

 

 

写起来太多了，希望没有写错了。，基本和代码实现对应的

 

  1 /* ***********************************************
  2 Author        :kuangbin
  3 Created Time  :2013/8/24 13:45:56
  4 File Name     :F:\2013ACM练习\比赛练习\2013通化邀请赛\1006.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 double k[100],a[100],b[100];
 21 vector<double>p;
 22 const double eps = 1e-8;
 23 void add(double a1,double b1,double c1)
 24 {
 25     if(fabs(a1) < eps && fabs(b1) < eps)
 26         return;
 27     if(fabs(a1) < eps)
 28     {
 29         double x = -c1/b1;
 30         if(x >= 0 && x <= 100)
 31             p.push_back(x);
 32         return;
 33     }
 34     double tmp = b1*b1 - 4*a1*c1;
 35     if(fabs(tmp) < eps)
 36     {
 37         double x = -b1/(2*a1);
 38         if(x >= 0 && x <= 100)
 39             p.push_back(x);
 40         return;
 41     }
 42     else if(tmp <=-eps)
 43     {
 44         return;
 45     }
 46     double x1 = (-b1+sqrt(tmp))/(2*a1);
 47     double x2 = (-b1-sqrt(tmp))/(2*a1);
 48     if(x1 >= 0 && x1 <= 100)
 49         p.push_back(x1);
 50     if(x2 >= 0 && x2 <= 100)
 51         p.push_back(x2);
 52 }
 53 double calc(double x)
 54 {
 55     return x*sqrt(1+x*x)/2 + log(x+sqrt(1+x*x))/2;
 56 }
 57 int main()
 58 {
 59     //freopen("in.txt","r",stdin);
 60     //freopen("out.txt","w",stdout);
 61     int T;
 62     int n;
 63     scanf("%d",&T);
 64     while(T--)
 65     {
 66         scanf("%d",&n);
 67         for(int i = 0;i < n;i++)
 68             scanf("%lf%lf%lf",&k[i],&a[i],&b[i]);
 69         p.clear();
 70         for(int i = 0;i < n;i++)
 71             add(k[i],-2*k[i]*a[i],k[i]*a[i]*a[i]+b[i]-100);
 72         for(int i = 0;i < n;i++)
 73             for(int j = i+1;j < n;j++)
 74             {
 75                 add(k[i]-k[j],2*k[j]*a[j]-2*k[i]*a[i],k[i]*a[i]*a[i]+b[i]-k[j]*a[j]*a[j]-b[j]);
 76             }
 77         p.push_back(0);
 78         p.push_back(100);
 79         double ans = 0;
 80         sort(p.begin(),p.end());
 81         int sz = p.size();
 82         for(int i = 1;i < sz;i++)
 83         {
 84             if(p[i] - p[i-1] < eps)continue;
 85             double tmp = (p[i] + p[i-1])/2;
 86             int tt = 0;
 87             for(int j = 0;j < n;j++)
 88                 if( k[j]*(tmp-a[j])*(tmp-a[j])+b[j] <   k[tt]*(tmp-a[tt])*(tmp-a[tt])+b[tt])
 89                     tt = j;
 90             if(k[tt]*(tmp-a[tt])*(tmp-a[tt])+b[tt] > 100)
 91             {
 92                 ans += p[i] - p[i-1];
 93                 continue;
 94             }
 95             ans += (calc(2*k[tt]*(p[i]-a[tt]))-calc(2*k[tt]*(p[i-1]-a[tt])))/(2*k[tt]);
 96 
 97         }
 98         printf("%.2lf\n",ans);
 99     }
100     return 0;
101 }

 

 

 

 

 

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