uva 1421 & uvalive 4253 Archery（二分）

题目连接：uva 1421 & uvalive 4253 Archery

题目大意：有n个平行与x轴的线段，每条线段代表一个靶子。你的任务是判断是否可以站在x轴上[0,w]区间的某个位置射箭，使得箭能穿过所有靶子。

解题思路：二分人站的位置，对于每个位置，维护靶子的可以击中的角度，如果出现靶子的区间,l > R,人左移；r < L人右移。

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <algorithm>
using namespace std;
#define max(a,b) (a)>(b)?(a):(b)
#define min(a,b) (a)<(b)?(a):(b)


const int N = 5005;
const double esp = 1e-6;

int n, t;
double w;

struct Seg {
	double d, l, r;
}s[N];

bool cmp(Seg a, Seg b) {
	return a.d < b.d;
}

int judge(double x) {
	double L = atan2(s[0].d, s[0].r - x);  
	double R = atan2(s[0].d, s[0].l - x);  
	for (int i = 1; i < n; i++) {  
		double l = atan2(s[i].d, s[i].r - x);  
		double r = atan2(s[i].d, s[i].l - x);  
		if (r - L < -esp)  
			return -1;  
		if (l - R > esp)  
			return 1;  
		L = max(L, l);  
		R = min(R, r);  
	}  
	return 0;
}

bool solve () {
	sort(s, s + n, cmp);
	double l = 0, r = w;
	while (r - l > esp) {
		double mid = (l+r)/2;
		int k = judge(mid);
		if (k == 0) return true;
		else if (k == -1) r = mid;
		else l = mid;
	}
	return false;
}

int main () {
	scanf("%d", &t);
	while (t--) {
		scanf("%lf%d", &w, &n);  
		for (int i = 0; i < n; i++)  
			scanf("%lf%lf%lf", &s[i].d, &s[i].l, &s[i].r);  
		printf("%s\n", solve() ? "YES" : "NO");
	}
	return 0;
}