UVA1356 Bridge (微积分 + 自适应辛普森法)

关键之处就在与一句话:

在(a, b)上可导的函数f(x)的弧长就g(x) = sqrt(1 + f'(x) ^ 2)在(a, b)上的定积分

所以说，题中的弧线明显为抛物线，那么它的长度就很可做了

然而你求的是深度……

没关系，假设深度depth和弧长L有一个函数L = h(depth)，这个函数明显是单调增的

所以就可以二分答案

code：

#include <bits/stdc++.h>
#define EPS 1e-5
using namespace std;

template <typename T> inline void read(T &x) {
    int c = getchar();
    bool f = false;
    for (x = 0; !isdigit(c); c = getchar()) {
        if (c == '-') {
            f = true;
        }
    }
    for (; isdigit(c); c = getchar()) {
        x = x * 10 + c - '0';
    }
    if (f) {
        x = -x;
    }
}

#define Debug(...) fprintf(stderr, __VA_ARGS__)
//#define LOCAL

double A;

double F(double x) {
    //the initial function
    return sqrt(1 + 4 * A * A * x * x); 
}

double calc(double a, double b) {
    double c = a + (b - a) / 2;
    return (F(a) + 4 * F(c) + F(b)) * (b - a) / 6.0000;
}

double asr(double a, double b, double c, double eps, double Border) {
    double L = calc(a, c), R = calc(c, b);
    if(fabs(L + R - Border) <= 15 * eps) return L + R + (L + R - Border) / 15.0000;
    return asr(a, c, a + (c - a) / 2, eps / 2, L) + asr(c, b, (b - c) / 2 + c, eps / 2, R);
}

double Simpson(double a, double b, double eps) {
    return asr(a, b, a + (b - a) / 2, eps, calc(a, b));
}

int T, D, H, B, L;

signed main() {
    read(T);
    for(int i = 1; i <= T; i++) {
        read(D), read(H), read(B), read(L);
        int n = (B + D - 1) / D;
        double stdD = (double) B * 1.000 / n;
        double stdL = (double) L * 1.000 / n;
        double l = 0.00000, r = (double) H;
        while(r - l > EPS) {
            double mid = (r - l) / 2 + l;
            A = 4 * mid / stdD / stdD;
            if(Simpson(0, stdD / 2, 1e-5) * 2 < stdL) l = mid;
            else r = mid;
        }
        printf("Case %d:\n%.2lf\n", i, H - l);
        if(i != T) puts("");
    }
#ifdef LOCAL
    Debug("My Time: %.3lfms\n", (double)clock() / CLOCKS_PER_SEC);
#endif
    return 0;
}

UVA1356 Bridge (微积分 + 自适应辛普森法)

猜你喜欢