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UVa 204 Robot Crash

news 2025/10/9 6:03:11

题目理解

本题要求判断两个机器人是否会在水平条带区域内发生碰撞。两个机器人从条带两端同时发射，在遇到墙壁时会按照反射定律反弹。我们需要找到它们是否会在同一位置、同一时间（时间差不超过0.5秒）相遇。

关键点分析

坐标系与运动范围：条带高度为10单位（0 ≤ y ≤ 10），机器人从左右两端发射
角度限制：
- 左枪角度：-85° 到 85°（相对于正x轴逆时针）
- 右枪角度：95° 到 180° 或 -95° 到 -180°
碰撞条件：在同一位置且时间差不超过0.5秒
墙壁反射：入射角等于反射角，速度大小不变

解题思路

核心算法

坐标变换与镜像处理：
- 将右枪机器人的运动镜像到左枪坐标系中考虑
- 通过交换坐标和速度方向来统一处理
路径构建：
- 根据初始位置、速度和运动时间构建机器人的运动轨迹
- 考虑墙壁反射，将轨迹分段处理
碰撞检测：
- 计算两机器人可能相遇的时间窗口
- 检查在这段时间内它们的轨迹段是否有交点
- 考虑时间差不超过0.5秒的限制
边界处理：
- 使用镜像法处理y坐标超出[0,10]范围的情况
- 确保轨迹计算在有效范围内

算法复杂度

时间复杂度：O(k)，其中k是轨迹段的数量
空间复杂度：O(k)，用于存储轨迹段

AC代码

// Robot Crash
// UVa ID: 204
// Verdict: Accepted
// Submission Date: 2025-10-08
// UVa Run Time: 0.010s
//
// 版权所有（C）2025，何昊。metaphysis # yeah dot net#include <bits/stdc++.h>
using namespace std;const double PI = acos(-1.0);                    // 圆周率
const double DEG_TO_RAD = PI / 180.0;            // 角度转弧度系数
const double EPS = 1e-7;                         // 浮点数比较精度
const int LENGTH = (1 << 20);                    // 输入缓冲区大小// 快速读取下一个字符
inline int nextChar() {static char buffer[LENGTH], *p = buffer, *end = buffer;if (p == end) {if ((end = buffer + fread(buffer, 1, LENGTH, stdin)) == buffer) return EOF;p = buffer;}return *p++;
}// 快速读取双精度浮点数
inline double readDouble() {int ch, sign = 1;double value = 0, decimal = 0, div = 1;while (isspace(ch = nextChar()));if (ch == '-') {sign = -1;ch = nextChar();} else if (ch == '+') {ch = nextChar();}while (isdigit(ch)) {value = value * 10 + (ch - '0');ch = nextChar();}if (ch == '.') {ch = nextChar();while (isdigit(ch)) {decimal = decimal * 10 + (ch - '0');div *= 10;ch = nextChar();}}return sign * (value + decimal / div);
}// 浮点数相等比较
inline bool eq(double a, double b) { return fabs(a - b) < EPS; }
// 浮点数小于等于比较
inline bool le(double a, double b) { return a < b + EPS; }
// 浮点数大于等于比较
inline bool ge(double a, double b) { return a + EPS > b; }// 点结构体
struct Point {double x, y;Point(double x = 0, double y = 0) : x(x), y(y) {}// 点比较：先按x坐标，再按y坐标bool operator<(const Point& p) const {return eq(x, p.x) ? (y < p.y) : (x < p.x);}
};// 向量结构体
struct Vector {double x, y;Vector(double x = 0, double y = 0) : x(x), y(y) {}
};// 线段结构体
struct Segment {Point start, end;Segment() : start(Point(0,0)), end(Point(0,0)) {}Segment(Point s, Point e) : start(s), end(e) {}
};// 直线结构体：ax + by + c = 0
struct Line {double a, b, c;Line(double a = 0, double b = 0, double c = 0) : a(a), b(b), c(c) {}// 通过两点构造直线Line(Point p1, Point p2) {a = p1.y - p2.y;b = p2.x - p1.x;c = -a * p1.x - b * p1.y;}// 判断点是否在直线上bool contains(Point p) const {return eq(a * p.x + b * p.y + c, 0);}
};// 检查两个区间是否重叠
inline bool intervalsOverlap(double a1, double a2, double b1, double b2) {if (a1 > a2) swap(a1, a2);if (b1 > b2) swap(b1, b2);return le(max(a1, b1), min(a2, b2));
}// 判断两线段是否相交，如果相交则返回交点
bool segmentsIntersect(const Segment& s1, const Segment& s2, Segment* result) {// 检查包围盒是否重叠if (!intervalsOverlap(s1.start.x, s1.end.x, s2.start.x, s2.end.x) ||!intervalsOverlap(s1.start.y, s1.end.y, s2.start.y, s2.end.y)) {return false;}Line line1(s1.start, s1.end);Line line2(s2.start, s2.end);double det = line1.a * line2.b - line2.a * line1.b;// 处理平行或共线情况if (eq(det, 0)) {if (!line1.contains(s2.start) || !line2.contains(s1.start)) return false;Point p1 = s1.start, p2 = s1.end;Point q1 = s2.start, q2 = s2.end;if (p2 < p1) swap(p1, p2);if (q2 < q1) swap(q1, q2);result->start = max(p1, q1);result->end = min(p2, q2);return true;}// 计算交点坐标double x = (line2.c * line1.b - line1.c * line2.b) / det;double y = (line2.a * line1.c - line1.a * line2.c) / det;result->start = Point(x, y);result->end = Point(x, y);// 检查交点是否在线段范围内return le(min(s1.start.x, s1.end.x), x) && le(x, max(s1.start.x, s1.end.x)) &&le(min(s1.start.y, s1.end.y), y) && le(y, max(s1.start.y, s1.end.y)) &&le(min(s2.start.x, s2.end.x), x) && le(x, max(s2.start.x, s2.end.x)) &&le(min(s2.start.y, s2.end.y), y) && le(y, max(s2.start.y, s2.end.y));
}// 调整点坐标到边界范围内，并相应调整速度向量
void adjustToBounds(Point& p, Vector& v) {if (p.y > 10) {double cycles = floor(p.y * 0.05 + EPS);p.y -= cycles * 20;if (p.y > 10) {p.y = 20 - p.y;v.y = -v.y;}} else if (p.y < 0) {double cycles = floor(-p.y * 0.05 + EPS);p.y += cycles * 20;if (p.y < -10) {p.y += 20;} else {p.y = -p.y;v.y = -v.y;}}
}// 构建机器人在给定时间范围内的运动路径（考虑墙壁反射）
vector<Segment> buildPath(Point pos, Vector vel, double startTime, double endTime) {vector<Segment> path;// 调整起始位置pos.x += vel.x * startTime;pos.y += vel.y * startTime;adjustToBounds(pos, vel);double time = startTime;while (true) {double nextTime = endTime;// 计算下一次碰到墙壁的时间if (!eq(vel.y, 0)) {if (vel.y > 0) {nextTime = time + (10 - pos.y) / vel.y;} else {nextTime = time - pos.y / vel.y;}if (nextTime > endTime) nextTime = endTime;}// 计算下一位置Point nextPos(pos.x + vel.x * (nextTime - time), pos.y + vel.y * (nextTime - time));path.push_back(Segment(pos, nextPos));// 更新位置和时间pos = nextPos;time = nextTime;vel.y = -vel.y;  // 碰到墙壁，y方向速度反向if (ge(time, endTime)) break;}return path;
}// 检查特定情况下的碰撞
bool checkCollisionCase(const Point& p1, const Vector& v1, const Point& p2,const Vector& v2, bool flipped, double* bestTime, Point* collisionPoint) {double dist = p2.x - p1.x;if (eq(dist, 0)) {// 特殊情况：初始x坐标相同if (eq(p2.y, p1.y) && !flipped) {*bestTime = 0;*collisionPoint = p1;return true;}return false;}if (dist < 0) return false;// 计算相遇时间double meetTime = dist / (v1.x - v2.x);double rightMaxTime = meetTime + v1.x / (v1.x - v2.x) * 0.5;if (rightMaxTime > *bestTime) {rightMaxTime = *bestTime;if (!ge(rightMaxTime, meetTime)) return false;}double leftMinTime = max(0.0, meetTime + v2.x / (v1.x - v2.x) * 0.5);// 构建两机器人的运动路径vector<Segment> rightPath = buildPath(p2, v2, meetTime, rightMaxTime);vector<Segment> leftPath = buildPath(p1, v1, leftMinTime, meetTime);reverse(leftPath.begin(), leftPath.end());Segment intersection(Point(0,0), Point(0,0));bool found = false;// 检查路径段是否有交点for (const auto& seg2 : rightPath) {for (const auto& seg1 : leftPath) {if (segmentsIntersect(seg1, seg2, &intersection)) {found = true;break;}}if (found) break;}if (!found) return false;// 计算碰撞时间和位置double collisionTime = (intersection.end.x - p2.x) / v2.x;double x = flipped ? (p1.x + p2.x - intersection.end.x) : intersection.end.x;// 更新最佳碰撞点if (eq(*bestTime, collisionTime)) {if (x < collisionPoint->x) {collisionPoint->x = x;collisionPoint->y = intersection.end.y;}} else {*bestTime = collisionTime;collisionPoint->x = x;collisionPoint->y = intersection.end.y;}return true;
}// 主碰撞检测函数
bool checkCollision(Point p1, Vector v1, Point p2, Vector v2, Point* result) {double bestTime = numeric_limits<double>::max();bool found = checkCollisionCase(p1, v1, p2, v2, false, &bestTime, result);// 交换位置和速度，检查镜像情况swap(p1.y, p2.y); swap(v1, v2);v1.x = -v1.x; v2.x = -v2.x;found |= checkCollisionCase(p1, v1, p2, v2, true, &bestTime, result);return found;
}int main(int argc, char* argv[]) {cin.tie(0), cout.tie(0), ios::sync_with_stdio(false);int caseNum = 0;Point robot1, robot2, collisionPoint;double angle1, angle2, speed1, speed2;// 读取输入直到文件结束while (true) {robot1.x = readDouble(); robot1.y = readDouble();angle1 = readDouble(); speed1 = readDouble();robot2.x = readDouble(); robot2.y = readDouble();angle2 = readDouble(); speed2 = readDouble();// 检查结束条件if (robot1.x == 0 && robot1.y == 0 && angle1 == 0 && speed1 == 0 &&robot2.x == 0 && robot2.y == 0 && angle2 == 0 && speed2 == 0) break;// 角度转弧度angle1 *= DEG_TO_RAD;angle2 *= DEG_TO_RAD;// 计算速度向量Vector vel1(speed1 * cos(angle1), speed1 * sin(angle1));Vector vel2(speed2 * cos(angle2), speed2 * sin(angle2));// 检查碰撞并输出结果if (checkCollision(robot1, vel1, robot2, vel2, &collisionPoint)) {printf("Robot Problem #%d: Robots collide at (%.2lf,%.2lf)\n\n",++caseNum, collisionPoint.x + EPS, collisionPoint.y + EPS);} else {printf("Robot Problem #%d: Robots do not collide.\n\n", ++caseNum);}}return 0;
}