UVa Problem 10136 Chocolate Chip Cookies （巧克力片饼干）

// Chocolate Chip Cookies （巧克力片饼干）// PC/UVa IDs: 111304/10136, Popularity: C, Success rate: average Level: 3// Verdict: Accepted// Submission Date: 2011-11-05// UVa Run Time: 0.020s//// 版权所有（C）2011，邱秋。metaphysis # yeah dot net//// [解题方法]// 若两片巧克力距离小于等于模具的直径，则可以移动模具，使得两片巧克力同时处于模具的边界上，这样再// 确定处于这个模具内的巧克力片数量就变得相对容易了，由于要使得模具内的巧克力片最多，只需枚举所有// 这样的模具位置，求其模具内巧克力片的最大值即可。#include <iostream>#include <sstream>#include <cmath>using namespace std;#define MAXN 200struct point{double x, y;};point chips[MAXN], center, median;int totalChips;// 计算圆心在 center 处，半径为 2.5cm 的模具内有多少巧克力片。int chipsInCutter(){int nCount = 0;for (int i = 0; i < totalChips; i++){double distance = sqrt(pow(chips[i].x - center.x, 2) +pow(chips[i].y - center.y, 2));if (distance <= 2.50)nCount++;}return nCount;}int main(int ac, char *av[]){istringstream iss;string line;int cases;bool printEmptyLine = false;cin >> cases;cin.ignore();getline(cin, line);while (cases--){// 读取巧克力片的位置。totalChips = 0;while (getline(cin, line), line.length()){iss.clear();iss.str(line);iss >> chips[totalChips].x >> chips[totalChips].y;totalChips++;}// 若两点距离小于等于模具的直径，则可求出模具的圆心位置，使得两点刚好在模具的// 边界上，注意，如果任意两点的距离都大于 5.0 cm，则最大巧克力片数为 1。int maxChips = 1;for (int i = 0; i < totalChips - 1; i++)for (int j = i + 1; j < totalChips; j++){double distance = sqrt(pow(chips[i].x - chips[j].x, 2) + pow(chips[i].y - chips[j].y, 2));if (distance > 5.0)continue;// 先处理特殊情况。if (chips[i].x == chips[j].x){center.x = chips[i].x + sqrt(2.50 * 2.50 -pow(fabs(chips[i].y - chips[j].y) / 2.0, 2));center.y = (chips[i].y + chips[j].y) / 2.0;maxChips = max(maxChips, chipsInCutter());center.x = chips[i].x - sqrt(2.50 * 2.50 -pow((chips[i].y - chips[j].y) / 2.0, 2));maxChips = max(maxChips, chipsInCutter());continue;}if (chips[i].y == chips[j].y){center.x = (chips[i].x + chips[j].x) / 2.0;center.y = chips[i].y + sqrt(2.50 * 2.50 -pow(fabs(chips[i].x - chips[j].x) / 2.0, 2));maxChips = max(maxChips, chipsInCutter());center.y = chips[i].y - sqrt(2.50 * 2.50 -pow(fabs(chips[i].x - chips[j].x) / 2.0, 2));maxChips = max(maxChips, chipsInCutter());continue;}// 根据经过圆心 center 的直线与过点 i，j 的直线垂直的关// 系计算圆心的坐标。先求出点 i，j 连线的中点。注意会有两// 个圆满足条件（若两点距离为 5.0cm，则两圆重合）。point median;median.x = (chips[i].x + chips[j].x) / 2.0;median.y = (chips[i].y + chips[j].y) / 2.0;// 求点 i，j 连线中点至圆心的距离，进而求圆心坐标。double slope = -(chips[j].x - chips[i].x) /(chips[j].y - chips[i].y);double segment = sqrt(2.50 * 2.50 -(pow(chips[i].x - chips[j].x, 2) +pow(chips[i].y - chips[j].y, 2)) / 4.0);double A = atan(slope);center.x = median.x + segment * cos(A);center.y = median.y + segment * sin(A);maxChips = max(maxChips, chipsInCutter());center.x = median.x - segment * cos(A);center.y = median.y - segment * sin(A);maxChips = max(maxChips, chipsInCutter());}if (printEmptyLine)cout << endl;elseprintEmptyLine = true;cout << maxChips << endl;}return 0;}