[codeforces 241]C. Mirror Box

[codeforces 241]C. Mirror Box

试题描述

Mirror Box is a name of a popular game in the Iranian National Amusement Park (INAP). There is a wooden box, 10⁵ cm long and 100cm high in this game. Some parts of the box's ceiling and floor are covered by mirrors. There are two negligibly small holes in the opposite sides of the box at heights h_l and h_r centimeters above the floor. The picture below shows what the box looks like.

In the game, you will be given a laser gun to shoot once. The laser beam must enter from one hole and exit from the other one. Each mirror has a preset number v_i, which shows the number of points players gain if their laser beam hits that mirror. Also — to make things even funnier — the beam must not hit any mirror more than once.

Given the information about the box, your task is to find the maximum score a player may gain. Please note that the reflection obeys the law "the angle of incidence equals the angle of reflection".

输入

The first line of the input contains three space-separated integers h_l, h_r, n (0 < h_l, h_r < 100, 0 ≤ n ≤ 100) — the heights of the holes and the number of the mirrors.

Next n lines contain the descriptions of the mirrors. The i-th line contains space-separated v_i, c_i, a_i, b_i; the integer v_i (1 ≤ v_i ≤ 1000) is the score for the i-th mirror; the character ci denotes i-th mirror's position — the mirror is on the ceiling if c_i equals "T" and on the floor if c_i equals "F"; integers a_i and b_i (0 ≤ a_i < b_i ≤ 10⁵) represent the x-coordinates of the beginning and the end of the mirror.

No two mirrors will share a common point. Consider that the x coordinate increases in the direction from left to right, so the border with the hole at height h_l has the x coordinate equal to 0 and the border with the hole at height h_r has the x coordinate equal to 10⁵.

输出

The only line of output should contain a single integer — the maximum possible score a player could gain.

输入示例

 T
 F
 T
 F
 F
 F
 F
 T
 T  

输出示例

数据规模及约定

见“输入”

题解

因为光线不能两次触碰到同一面镜子，所以拐点不会超过 n 个，我们不妨枚举拐点个数。确定了拐点个数后，我们设 t 是一个完整斜边（从底边反射上来再触碰到顶）在底边上投影的距离，不难列出一个一元一次方程，把 t 解出来后再遍历一遍每个镜子检查答案是否合法，合法就更新最优得分。枚举拐点个数 O(n)，检查合法性 O(n)，总时间复杂度 O(n²).

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <stack>
#include <vector>
#include <queue>
#include <cstring>
#include <string>
#include <map>
#include <set>
using namespace std;

const int BufferSize = 1 << 16;
char buffer[BufferSize], *Head, *Tail;
inline char Getchar() {
    if(Head == Tail) {
        int l = fread(buffer, 1, BufferSize, stdin);
        Tail = (Head = buffer) + l;
    }
    return *Head++;
}
int read() {
    int x = 0, f = 1; char c = getchar();
    while(!isdigit(c)){ if(c == '-') f = -1; c = getchar(); }
    while(isdigit(c)){ x = x * 10 + c - '0'; c = getchar(); }
    return x * f;
}

#define maxn 110
#define len 100000
const double eps = 1e-6;
int h1, h2, n, cu, cd, ans;
struct Mirror {
	int v, l, r;
	bool operator < (const Mirror& t) const { return l < t.l; }
} up[maxn], down[maxn];

void check(bool cur, double t, double st) {
	bool ok = 0;
	int lu = 1, lastu = 0, ld = 1, lastd = 0, tmp = 0;
	for(; st - len < eps && (lu <= cu || ld <= cd);) {
		if(cur) {
			while(lu < cu && st - up[lu].r > eps) lu++;
			if(st - up[lu].l < eps || st - up[lu].r > eps || lu == lastu) return ;
			tmp += up[lu].v; lastu = lu;
		}
		else {
			while(ld < cd && st - down[ld].r > eps) ld++;
			if(st - down[ld].l < eps || st - down[ld].r > eps || ld == lastd) return ;
			tmp += down[ld].v; lastd = ld;
		}
		st += t; cur ^= 1;
	}
	ans = max(ans, tmp);
	return ;
}

int main() {
	h1 = read(); h2 = read(); n = read();
	for(int i = 1; i <= n; i++) {
		int v = read();
		char tp = getchar();
		while(!isalpha(tp)) tp = getchar();
		if(tp == 'T') up[++cu].v = v, up[cu].l = read(), up[cu].r = read();
		if(tp == 'F') down[++cd].v = v, down[cd].l = read(), down[cd].r = read();
	}

	sort(up + 1, up + cu + 1);
	sort(down + 1, down + cd + 1);
	for(int k = 0; k <= n; k++) {
		// down 0; up 1;
		double t;
		if(k & 1) {
			t = (double)len / ((double)h1 / 100.0 + (100.0 - h2) / 100.0 + k);
			check(0, t, t * h1 / 100.0);
			t = (double)len / ((100.0 - h1) / 100.0 + (double)h2 / 100.0 + k);
			check(1, t, t * (100.0 - h1) / 100.0);
		}
		else {
			t = (double)len / ((double)h1 / 100.0 + (double)h2 / 100.0 + k);
			check(0, t, t * h1 / 100.0);
			t = (double)len / ((100.0 - h1) / 100.0 + (100.0 - h2) / 100.0 + k);
			check(1, t, t * (100.0 - h1) / 100.0);
		}
	}

	printf("%d\n", ans);

	return 0;
}