POJ 1151 Atlantis（离散化+扫描线）

题目大意：

求N个矩形的并的总面积。http://poj.org/problem?id=1151

参考了黑书上讲离散化的那两页，想到了上学期学的计算机图形学里的多边形填充算法。

src:

/********************************************************************created:2013/03/28created:28:3:2013   11:43filename: H:\PE\USA\POJ.1151.Atlantis.cppfile path:H:\PE\USAfile base:POJ.1151.Atlantisfile ext:cppauthor:Justme0 (http://blog.csdn.net/Justme0)purpose:http://poj.org/problem?id=1151求矩形的并的总面积*********************************************************************/// #define ONLINE_JUDGE#define _CRT_SECURE_NO_WARNINGS#include <iostream>#include <algorithm>#include <vector>#include <string>#include <cassert>using namespace std;// 水平线段struct Horizon {double xL;// 左端点的横坐标double xR;// 右端点的横坐标double y;// 水平线的纵坐标bool isUp;// 矩形的上边or下边Horizon() : xL(0), xR(0), y(0), isUp(false) {}Horizon(double _xL, double _xR, double _y, bool _isUp): xL(_xL),xR(_xR),y(_y),isUp(_isUp) {}bool operator<(const Horizon &other) const {return this->y < other.y;}};// 返回：输入是否结束bool input(vector<double> &xVec, vector<Horizon> &hrzVec) {int cnt;cin >> cnt;if (0 == cnt) {return false;}double Ax, Ay, Bx, By;while (cnt--) {cin >> Ax >> Ay >> Bx >> By;xVec.push_back(Ax);xVec.push_back(Bx);hrzVec.push_back(Horizon(Ax, Bx, Ay, false));hrzVec.push_back(Horizon(Ax, Bx, By, true));}return true;}void solve(vector<double> &xVec, const vector<Horizon> &hrzVec, double &result) {result = 0;sort(xVec.begin(), xVec.end());xVec.erase(unique(xVec.begin(), xVec.end()), xVec.end());sort(hrzVec.begin(), hrzVec.end());for (vector<double>::size_type i = 0; i < xVec.size() - 1; ++i) {double L = xVec[i];double R = xVec[i + 1];double width = R - L;assert(width > 0);// 等于0的已uniqueint count = 0;// 计数器，正数表示被覆盖，0表示未被覆盖（有count个矩形覆盖了该区域）Horizon former;for (vector<Horizon>::size_type j = 0; j < hrzVec.size(); ++j) {Horizon current = hrzVec[j];if (current.xL <= L && R <= current.xR) {assert(count >= 0);if (count > 0) {double height = current.y - former.y;assert(height >= 0);result += width * height;}former = current;current.isUp ? --count : ++count;assert(count >= 0);}}assert(count == 0);}}void output( int testCase, double result ) {printf("Test case #%d\nTotal explored area: %.2f\n\n", testCase, result);}int main(){#ifndef ONLINE_JUDGEfreopen("cin.txt", "r", stdin);#endiffor (int testCase = 1; ; ++testCase) {vector<double> xVec;vector<Horizon> hrzVec;if (!input( xVec, hrzVec)) {break;}double result = 0;solve(xVec, hrzVec, result);output(testCase, result);}return 0;}

在网上看到有的ACMER把程序分为输入、解决问题、输出的三层结构，觉得挺好的，打算以后也这么做，免得全挤在一个main里，而且这样也便于调试。我尽量不用全局变量，这样各函数调用时消息的传递就很明确了。