zoj 2369 Two Cylinders

zoj 2369 Two Cylinders

链接：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=2369

题意：已知两个无限长的圆柱的半径，它们轴线相交且垂直，求相交部分的体积。

思路：以相交点为坐标原点，建立三维坐标系，列出 两个圆柱的公式，把三维转化为一维积分公式

　　　　V₁ : x² + y² = r₁²　　　　==>  y = sqrt ( r₁² - x²  );

　　　　V₂ : x² + z² = r₂²　　　　==>  z = sqrt ( r₂² - x²  );

　　　　V = ∫∫∫ dv = ∫∫∫ dxdydz = ∫(∫∫ dydz)dx　　　　积分区间：( 0 <= x <= r₁ ,  0 <= x <= r₂  )

　　　　所以： V = ∫₀^{min(r₁, r₂)} sqrt(( r₁²-x²) * (r₂²-x²));

计算这个积分，我用了好几种方法，但是只有 变步长simpon 积分达到精度要求，我也不清楚什么原因。下面是几种求积分的算法：

测试数据：

10
1 1
2 1
3 2
5 2
8 3
8 4
99 2
99 56
84 31
91 17

正确答案：

5.3333
12.1603
70.9361
123.0975
444.2909
778.2605
2488.0144
1869197.4256
498415.2860
164517.4621

变步长 simpon 积分：（得到正确结果）

 #include <iostream>
 #include <stdio.h>
 #include <algorithm>
 #include <queue>
 #include <map>
 #include <set>
 #include <vector>
 #include <cstring>
 #include <math.h>
 using namespace std;

 const double eps = 1e-;
 double r1, r2;

 double f(double x)
 {
     return . * sqrt((r1*r1 - x*x) * (r2*r2 - x*x));
 }

 double simpson(double a, double b)
 {
     int n, k;
     double h, t1, t2, s1, s2, ep, p, x;
     n = , h = b - a;

     t1 = h * ( f(a) + f(b) ) / .;        //计算T1 = (b-a)/2*[f(a) + f(b)]
     s1 = t1;        //T1代替S1
     ep = eps + .;
     while(ep >= eps)
     {
         p = ;
         for(k = ; k <= n-; k++)
         {
             x = a + (k + 0.5) * h;
             p += f(x);
         }
         t2 = (t1 + h * p) / .;    //计算T2n = Tn/2 + h/2 * [(累加)f(xk + 0.5)];
         s2 = (. * t2 - t1) / .; //计算Sn = (4*Tn - Tn) / 3;
         ep = fabs(s2 - s1);  //计算精度
         t1 = t2, s1 = s2, n += n, h /= .;
     }
     return s2;
 }

 int main()
 {
     int CASE;
     freopen("zoj2369.txt", "r", stdin);
     cin >> CASE;
     while(CASE--)
     {
         scanf("%lf%lf", &r1, &r2);
         if(r2 < r1)    swap(r1, r2);
         double t = simpson(., r1);
         printf("%.4lf\n", t);

     }
     return ;
 }

 /* 结果：
 5.3333
 12.1603
 70.9361
 123.0974
 444.2908
 778.2604
 2488.0144
 1869197.4256
 498415.2860
 164517.4621
 */

勒让德-高斯求积（精度不够）：

 #include <iostream>
 #include <stdio.h>
 #include <algorithm>
 #include <queue>
 #include <map>
 #include <set>
 #include <vector>
 #include <cstring>
 #include <math.h>
 using namespace std;

 const double eps = 1e-;
 double r1, r2;

 double f(double x)
 {
     return . * sqrt((r1*r1 - x*x) * (r2*r2 - x*x));
 }

 double lrgs(double a, double b)
 {
     int m, i, j;
     double s, p, ep, aa, bb, w, x, g, h;
     static double t[] = {-0.9061798459, -0.5384693101, 0.0, 0.5384693101, 0.9061798459};
     static double c[] = {0.2369268851, 0.4786286705, 0.5688888889, 0.4786286705, 0.2369268851};
     m = ;
     h = b - a; s = fabs(0.001 * h);
     p = 1.0e+35, ep = eps + 1.0;
     while(ep >= eps && fabs(h) > s)
     {
         g = .;
         for(i = ; i <= m; i++)
         {
             aa = a + (i-.) * h,  bb = a + i * h;
             w = .;
             for(j = ; j < ; j++)
             {
                 x = ((bb - aa) * t[j] + (bb + aa)) / .;
                 w += f(x) * c[j];
             }
             g += w;
         }
         g = g * h / .;
         ep = fabs(g-p) / (. + fabs(g));
         p = g, m += , h = (b-a) / m;
     }
     return g;
 }

 int main()
 {
     int t;
     freopen("zoj2369.txt", "r", stdin);
     scanf("%d", &t);
     while(t--)
     {
         scanf("%lf%lf", &r1, &r2);
         if(r1 > r2)    swap(r1, r2);
         printf("%.4lf\n", lrgs(, r1));
     }
     return ;
 }

 /*结果
 5.3333
 12.1619
 70.9440
 123.1138
 444.3504
 778.3592
 2488.3679
 1869425.2091
 498482.1853
 164540.5195
 */

龙贝格求积法（精度不够）：

 #include <iostream>
 #include <stdio.h>
 #include <algorithm>
 #include <vector>
 #include <cstring>
 #include <math.h>
 using namespace std;

 const double eps = 1e-;
 double r1, r2;

 double f(double x)
 {
     return . * sqrt((r1*r1 - x*x) * (r2*r2 - x*x));
 }

 double romb(double a, double b)
 {
     int n, m, i, k;
     double y[], h, ep, p, x, s, q;
     h = b - a;
     y[] = h * ((f(a) + f(b))) / .;
     m = , n = , ep = eps + 1.0;
     while(ep >= eps )
     {
         p = 0.0;
         for(i = ; i < n; i++)
         {
             x = a + (i+0.5) * h;
             p += f(x);
         }
         p = (y[] + h * p) /.;
         s = 1.0;
         for(k = ; k <= m; k++)
         {
             s = . * s;
             q = (s*p - y[k-]) / (s - .);
             y[k - ] = p, p = q;
         }
         ep = fabs(q - y[m-]);
         m = m + , y[m - ] = q, n += n, h /= .;
     }
     return q;
 }

 int main()
 {
     int t;
     freopen("zoj2369.txt", "r", stdin);
     freopen("zzin1.txt", "w", stdout);
     scanf("%d", &t);
     while(t--)
     {
         scanf("%lf%lf", &r1, &r2);
         if(r1 > r2)    swap(r1, r2);
         printf("%.4lf\n", romb(, r1));
     }
     return ;
 }

 /*结果
 5.3333
 12.1529
 70.8978
 123.0697
 444.1897
 778.0925
 2487.8023
 1869191.3654
 498410.2627
 164515.7323
 */

也不知最后两个代码那里写搓了，真心无力。

下面给出标程：

 #include <cmath>
 #include <cstdio>
 #include <algorithm>

 using namespace std;

 const long double H = 5e-;

 struct F {
     long double rr1, rr2;
     F(long double r1, long double r2) : rr1(r1 * r1), rr2(r2 * r2) {
     }
     long double operator()(long double x) const {
         x = x * x;
         return sqrt((rr2 - x) * (rr1 - x));
     }
 };

 int main() {
     int re;

     scanf("%d", &re);
     for (int ri = ; ri <= re; ++ri) {
         long double r1, r2;
         scanf("%Lf%Lf", &r1, &r2);
         if (r1 > r2) {
             swap(r1, r2);
         }
         int n = int(r1 / H);
         // int n = 1000000;
         long double h = r1 / n / ;
         F f(r1, r2);
         long double s = ;
         s += f() + f(r1);
         for (int i = ; i <  * n; ++i) {
             if (i & ) {
                 s +=  * f(i * h);
             } else {
                 s +=  * f(i * h);
             }
         }
         s /=  * n;
         s *=  * r1;
         printf("%.6Lf\n", s);
     }

     return ;
 }

网上看的大牛的代码，0ms秒过，我的用了290ms， 标程用了310ms

 #include <cstdio>
 #include <cmath>
 #include <algorithm>
 using namespace std;

 #define db double
 #define rt return
 #define cs const

 cs db eps = 1e-;
 inline int sig(db x){rt (x>eps)-(x<-eps);}

 db r1, r2;

 inline db f(db x) {
     rt . * sqrt(r1*r1-x*x) * sqrt(r2*r2-x*x);
 }

 inline db simpson(db a, db b) {
     rt (b-a)*(f(a) + .*f((a+b)/.) + f(b)) / .;
 }

 inline db calc(db a, db b, int depth) {
     db total = simpson(a, b), mid = (a+b) / .;
     db tmp = simpson(a, mid) + simpson(mid, b);
     if(sig(total-tmp) ==  && depth > ) rt total;
     rt calc(a, mid, depth + ) + calc(mid, b, depth + );
 }

 int main() {
 //    freopen("std.in", "r", stdin);
     int T;
     freopen("zoj2369.txt", "r", stdin);
     scanf("%d", &T);
     while(T--) {
         scanf("%lf%lf", &r1, &r2);
         if(sig(r1-r2) > ) swap(r1, r2);
         printf("%.4lf\n", . * calc(., r1, ));
     }
     rt ;

 }