ZOJ1104-Leaps Tall Buildings

Leaps Tall Buildings

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Time Limit: 2 Seconds      Memory Limit: 65536 KB      Special Judge

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It's a bird! It's a plane! It's coming right at us!

Although it sometimes seems like it, Superman can't fly (without a plane). Instead, he makes super-human leaps, especially over tall buildings. Since he never knows when he will need to catch a criminal, he can't register flight paths. To avoid hitting planes, he tries to keep his jumps as low to the ground as he can. Given a city-scape as input, find the angle and velocity of Superman's jump that minimizes his maximum altitude.

Recall that gravity provides an acceleration of 9.8 m/s² downwards and the formula for Superman's vertical distance from his starting location is d(t)=v t + 0.5 a t² where v is his initial velocity, a is his acceleration and t is time in seconds since the start of the leap.

Input:

Input consists of a sequence of city-scapes, each of the form

n
0 d₁
h₂ d₂
:
h_(n-1) d_(n-1)
0 d_n

Superman starts at ground level and leaps d₁+...+d_n metres, landing at ground level and clearing all of the buildings at heights h₂ to h_(n-1), each with the given widths. n will be at most 100.

Output:

Output is the angle and initial velocity that minimizes the height that Superman attains, both appearing on the same line. The values should be given to two decimal places and be accurate within 0.01 degrees or m/s, as appropriate.

Sample Input:

3
0 5
10 5
0 5
5
0 10.5
20 11.5
25 10
10 15
0 7

Diagram for Second City-scape

 
(Not to scale.)

Sample Output:

71.57 15.65
67.07 27.16

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Source: University of Waterloo Local Contest 1998.06.06

题意：有n个建筑，有一个超人要跳过所有建筑，问最低的高度和初始速度角度

解题思路：二分高度

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>

using namespace std;

#define LL long long
const int INF = 0x3f3f3f3f;
const double pi = acos(-1.0);
const double eps = 1e-8;

double x[105], y[105];
int n, cnt;
double v, k;

bool check(double h)
{
	double v1 = sqrt(2 * 9.8*h);
	double v2 = x[cnt - 1] / (v1 / 4.9);
	v = sqrt(v1*v1 + v2*v2);
	k = 180 * acos(v2 / v) / pi;
	for (int i = 0; i<cnt; i++)
	{
		double xx;
		if (x[i] > (x[cnt - 1] / 2)) xx = x[cnt - 1] - x[i];
		else xx=x[i];
		double t = xx / v2;
		double hh = v1*t - 4.9*t*t;
		if (hh<y[i]) return 0;
	}
	return 1;
}

int main()
{

	while (~scanf("%d", &n))
	{
		double xx = cnt = 0, a, b;
		for (int i = 1; i <= n; i++)
		{
			scanf("%lf%lf", &a, &b);
			x[cnt] = xx; y[cnt++] = a;
			x[cnt] = xx += b; y[cnt++] = a;
		}
		double l = 0, r =1.0*INF;
		while (r-l>eps)
		{
			double mid = (l + r) / 2;
			if (check(mid)) r = mid;
			else l = mid;
		}
		check(r);
		printf("%lf %lf\n", k, v);
	}
	return 0;
}