hdu 4498 Function Curve（分段曲线积分）

Function Curve

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 249    Accepted Submission(s): 93

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

 

题意：求上函数的0~100区间的曲线段的长度

题解：求出所有函数交点x坐标，将函数变成分段函数，然后对每一段求曲线积分就行了。。。可是曲线积分有点难求（对于我）。。。所以simpson一下，year~

         注意交点貌似挺多的，至少大于333个。。。我开始就拼命RE这里，另外simpson也TLE了一下。。还是挺靠感觉调的

#include<stdio.h>#include<stdlib.h>#include<math.h>#define eps 1e-8double a[53],b[53],k[53],point[33333];int all,n,id;int cmp(const void *a,const void *b){    return *(double *)a<*(double *)b?-1:1;}double Function(double x,int id){    return k[id]*(x-a[id])*(x-a[id])+b[id];}double Func(double x){    return sqrt(1+(2*k[id]*(x-a[id]))*(2*k[id]*(x-a[id])));}double simpson(double l,double r){    double mid=(l+r)/2.0;    return (r-l)*(Func(l)+4*Func(mid)+Func(r))/6.0;}double cal(double l,double r,double all){    double mid=(l+r)/2.0;    double L=simpson(l,mid);    double R=simpson(mid,r);    if(fabs(L+R-all)<=eps) return all;    else return cal(l,mid,L)+cal(mid,r,R);}void get_point(){    int i,j;    double x1,x2,aa,bb,cc;    for(i=0;i<n;i++)    {        if(b[i]>100) continue;        x1=sqrt((100-b[i])/k[i])+a[i];        x2=-sqrt((100-b[i])/k[i])+a[i];        if(-eps<x1&&x1<100+eps) point[all++]=x1;        if(-eps<x2&&x2<100+eps) point[all++]=x2;    }    for(i=0;i<n;i++)    {        for(j=i+1;j<n;j++)        {            aa=k[i]-k[j];            bb=-2*a[i]*k[i]+2*a[j]*k[j];            cc=a[i]*a[i]*k[i]+b[i]-a[j]*a[j]*k[j]-b[j];            if(bb*bb-4*aa*cc<0) continue;            x1=(sqrt(bb*bb-4*aa*cc)-bb)/(2*aa);            x2=(-sqrt(bb*bb-4*aa*cc)-bb)/(2*aa);            if(0<x1&&x1<100) point[all++]=x1;            if(0<x2&&x2<100) point[all++]=x2;        }    }}int main(){    int t,i,j;    double left,right,mid,res;    //freopen("t.txt","r",stdin);    scanf("%d",&t);    while(t--)    {        scanf("%d",&n);        for(i=0;i<n;i++) scanf("%lf%lf%lf",k+i,a+i,b+i);        all=0,point[all++]=0,point[all++]=100;        get_point();        qsort(point,all,sizeof(point[0]),cmp);        for(res=i=0;i<all-1;i++)        {            left=point[i],right=point[i+1];            mid=(left+right)/2;            if(right-left<eps) continue;            id=0;            for(j=1;j<n;j++)            {                if(Function(mid,j)-Function(mid,id)<0) id=j;            }            if(Function(mid,id)-100<0)                res+=cal(left,right,simpson(left,right));            else res+=right-left;        }        printf("%.2lf\n",res);    }    return 0;}