UVa Problem 10136 Chocolate Chip Cookies （巧克力片餅乾）

// Chocolate Chip Cookies （巧克力片餅乾）// PC/UVa IDs: 111304/10136, Popularity: C, Success rate: average Level: 3// Verdict: Accepted// Submission Date: 2011-11-05// UVa Run Time: 0.020s//// 著作權（C）2011，邱秋。metaphysis # yeah dot net//// [解題方法]// 若兩片巧克力距離小於等於模具的直徑，則可以移動模具，使得兩片巧克力同時處於模具的邊界上，這樣再// 確定處於這個模具內的巧克力片數量就變得相對容易了，由於要使得模具內的巧克力片最多，只需枚舉所有// 這樣的模具位置，求其模具內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;// 計算圓心在 center 處，半徑為 2.5cm 的模具內有多少巧克力片。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--){// 讀取巧克力片的位置。totalChips = 0;while (getline(cin, line), line.length()){iss.clear();iss.str(line);iss >> chips[totalChips].x >> chips[totalChips].y;totalChips++;}// 若兩點距離小於等於模具的直徑，則可求出模具的圓心位置，使得兩點剛好在模具的// 邊界上，注意，如果任意兩點的距離都大於 5.0 cm，則最大巧克力片數為 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;// 先處理特殊情況。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;}// 根據經過圓心 center 的直線與過點 i，j 的直線垂直的關// 系計算圓心的座標。先求出點 i，j 連線的中點。注意會有兩// 個圓滿足條件（若兩點距離為 5.0cm，則兩圓重合）。point median;median.x = (chips[i].x + chips[j].x) / 2.0;median.y = (chips[i].y + chips[j].y) / 2.0;// 求點 i，j 連線中點至圓心的距離，進而求圓心座標。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;}