UVALive 6835 Space Golf（物理题）

题意:

n≤10个障碍物,一个球从起点弹到终点距离为d≤10000,反弹次数b≤15,重力g=1
不计中间过程的能量损失,问可以越过所有障碍物最小的初速度是多少？

分析:

由于反弹次数很小,我们可以考虑枚举反弹次数,由于障碍物个数也很少我们也可以枚举障碍物
假设所求的速度为v,vx=vcosθ,vy=vsinθ,根据水平时间与竖直时间相等可知,12lvx=vyg
可得l=2vxvyg=v2⋅2sinθcosθg=v2⋅sin2θg　①
对于相同的速度的情况下,θ=45度时,l取到最大值为最优,所以出射角应该≥45度
对于每个反弹次数,我们可以得到球运动的抛物线的宽度l=d/(i+1)
对于一个包含障碍物的抛物线周期,我们知道三个点(0,0),(x,hi),(l,0),其中x是所在抛物线周期的pi位置
根据抛物线两根式y=a(x−x1)(x−x2)=>y=ax(x−l),解得:a=hi/(x∗(x−l))
求导y=ax(x−l)得到抛物线的斜率方程y′=2ax−al,代入x=0得,tanθ=−al,即θ=atan(−al)　②
由①变形得,v2=lgsin2θ,再由②即可算出v2
问题解决,不要忘记特判碰到障碍物的情况

代码:

////  Created by TaoSama on 2015-12-07//  Copyright (c) 2015 TaoSama. All rights reserved.////#pragma comment(linker, "/STACK:1024000000,1024000000")#include <algorithm>#include <cctype>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iomanip>#include <iostream>#include <map>#include <queue>#include <string>#include <set>#include <vector>using namespace std;#define pr(x) cout << #x << " = " << x << "  "#define prln(x) cout << #x << " = " << x << endlconst int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;const double PI = acos(-1), EPS = 1e-8;int sgn(double x) {    return x < -EPS ? -1 : x < EPS ? 0 : 1;}int d, n, b;int p[15], h[15];double solve(int b) {    double l = 1.0 * d / (b + 1), maxv = 0;    for(int i = 1; i <= n; ++i) {        double x = fmod(p[i], l);        if(sgn(x) == 0) return 1e20; //collision        double a = h[i] / x / (x - l), k = -a * l;        double ang = max(atan2(k, 1), PI / 4); // >= 45;        maxv = max(maxv, l / sin(2 * ang));    }    return maxv;}int main() {#ifdef LOCAL    freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);//  freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);#endif    ios_base::sync_with_stdio(0);    while(scanf("%d%d%d", &d, &n, &b) == 3) {        for(int i = 1; i <= n; ++i) scanf("%d%d", p + i, h + i);        double ans = 1e20;        for(int i = 0; i <= b; ++i)            ans = min(ans, solve(i));        printf("%.5f\n", sqrt(ans));    }    return 0;}