HDU 4498 Function Curve （分段，算曲线积分）

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, … 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

 

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

 /* ***********************************************
 Author        :kuangbin
 Created Time  :2013/8/24 13:45:56
 File Name     :F:\2013ACM练习\比赛练习\2013通化邀请赛\1006.cpp
 ************************************************ */

 #include <stdio.h>
 #include <string.h>
 #include <iostream>
 #include <algorithm>
 #include <vector>
 #include <queue>
 #include <set>
 #include <map>
 #include <string>
 #include <math.h>
 #include <stdlib.h>
 #include <time.h>
 using namespace std;
 double k[],a[],b[];
 vector<double>p;
 const double eps = 1e-;
 void add(double a1,double b1,double c1)
 {
     if(fabs(a1) < eps && fabs(b1) < eps)
         return;
     if(fabs(a1) < eps)
     {
         double x = -c1/b1;
         if(x >=  && x <= )
             p.push_back(x);
         return;
     }
     double tmp = b1*b1 - *a1*c1;
     if(fabs(tmp) < eps)
     {
         double x = -b1/(*a1);
         if(x >=  && x <= )
             p.push_back(x);
         return;
     }
     else if(tmp <=-eps)
     {
         return;
     }
     double x1 = (-b1+sqrt(tmp))/(*a1);
     double x2 = (-b1-sqrt(tmp))/(*a1);
     if(x1 >=  && x1 <= )
         p.push_back(x1);
     if(x2 >=  && x2 <= )
         p.push_back(x2);
 }
 double calc(double x)
 {
     return x*sqrt(+x*x)/ + log(x+sqrt(+x*x))/;
 }
 int main()
 {
     //freopen("in.txt","r",stdin);
     //freopen("out.txt","w",stdout);
     int T;
     int n;
     scanf("%d",&T);
     while(T--)
     {
         scanf("%d",&n);
         for(int i = ;i < n;i++)
             scanf("%lf%lf%lf",&k[i],&a[i],&b[i]);
         p.clear();
         for(int i = ;i < n;i++)
             add(k[i],-*k[i]*a[i],k[i]*a[i]*a[i]+b[i]-);
         for(int i = ;i < n;i++)
             for(int j = i+;j < n;j++)
             {
                 add(k[i]-k[j],*k[j]*a[j]-*k[i]*a[i],k[i]*a[i]*a[i]+b[i]-k[j]*a[j]*a[j]-b[j]);
             }
         p.push_back();
         p.push_back();
         double ans = ;
         sort(p.begin(),p.end());
         int sz = p.size();
         for(int i = ;i < sz;i++)
         {
             if(p[i] - p[i-] < eps)continue;
             double tmp = (p[i] + p[i-])/;
             int tt = ;
             for(int j = ;j < n;j++)
                 if( k[j]*(tmp-a[j])*(tmp-a[j])+b[j] <   k[tt]*(tmp-a[tt])*(tmp-a[tt])+b[tt])
                     tt = j;
             if(k[tt]*(tmp-a[tt])*(tmp-a[tt])+b[tt] > )
             {
                 ans += p[i] - p[i-];
                 continue;
             }
             ans += (calc(*k[tt]*(p[i]-a[tt]))-calc(*k[tt]*(p[i-]-a[tt])))/(*k[tt]);

         }
         printf("%.2lf\n",ans);
     }
     return ;
 }