[BZOJ5099]Pionek

Description

给 (n) ((nle 2 imes 10 ^5)) 个向量，现在你在 ((0,0)) ，选择一些向量使你走的最远。

Solution

自己的想法：按极角排序后双指针 (l, r) 扫，若选择 (r + 1) 向量走的更远就 r++ ，否则 l++ ，用 ([l,r]) 的向量和与答案 (chkmax)。

这样是错的，虽然答案最后一定是一段连续的区间，但这个并不满足局部最优，所以可能 (r) 指针需要舍弃一些不优的右移而到一个更好的位置。

题解首先按极角排序，然后答案一定是某一个半平面，即选一根过原点的x轴，x正半轴的向量都选，也是双指针扫一遍即可。

code

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <fstream>

typedef long long LL;
typedef unsigned long long uLL;

#define SZ(x) ((int)x.size())
#define ALL(x) (x).begin(), (x).end()
#define MP(x, y) std::make_pair(x, y)
#define DEBUG(...) fprintf(stderr, __VA_ARGS__)
#define GO cerr << "GO" << endl;

using namespace std;

inline void proc_status()
{
	ifstream t("/proc/self/status");
	cerr << string(istreambuf_iterator<char>(t), istreambuf_iterator<char>()) << endl;
}

template<class T> inline T read() 
{
	register T x = 0; register int f = 1; register char c;
	while (!isdigit(c = getchar())) if (c == '-') f = -1;
	while (x = (x << 1) + (x << 3) + (c xor 48), isdigit(c = getchar()));
	return x * f;
}

template<typename T> inline bool chkmin(T &a, T b) { return a > b ? a = b, 1 : 0; }
template<typename T> inline bool chkmax(T &a, T b) { return a < b ? a = b, 1 : 0; }

const int maxN = (int) 2e5;
const double PI = acos(-1);

struct Vector
{
	int x, y;
	double angle;
	
	Vector(int x = 0, int y = 0) : x(x), y(y) { angle = atan2(y, x); }
	
	bool operator < (const Vector& B) const { return angle < B.angle; }

	Vector operator + (const Vector& B) const { return Vector(x + B.x, y + B.y); }
	Vector operator - (const Vector& B) const { return Vector(x - B.x, y - B.y); }
} ;

int n;
Vector vec[maxN * 2 + 2];

void Input()
{
	n = read<int>();
	for (int i = 1; i <= n; ++i)
	{
		int x = read<int>(), y = read<int>();
		vec[i] = Vector(x, y);
	}
}

void Init()
{
	sort(vec + 1, vec + 1 + n);
	for (int i = 1; i <= n; ++i) vec[i + n] = vec[i], vec[i + n].angle += 2 * PI;
	n <<= 1;
}

LL dis(Vector cur) { return (LL)cur.x * cur.x + (LL)cur.y * cur.y; }

void Solve()
{
	LL ans(0);
	Vector cur(0, 0);

	for (int i = 1, j = 1; j <= n / 2; ++j)
	{
		while (i < n + j and vec[i].angle - vec[j].angle < PI)
			cur = cur + vec[i++], chkmax(ans, dis(cur));
		cur = cur - vec[j];
		chkmax(ans, dis(cur));
	}

	printf("%lld
", ans);
}

int main() 
{
#ifndef ONLINE_JUDGE
	freopen("BZOJ5099.in", "r", stdin);
	freopen("BZOJ5099.out", "w", stdout);
#endif

	Input();

	Init();

	Solve();

	return 0;
}