Sicily 1004. I Conduit!

题目链接在此

参考的是这位大神的解题方法

这题要考虑的情况有很多。

另外两个double类型数据不能简单地以==来判断相等！

以下是源代码：

#include<iostream>#include<cmath>#include<algorithm>using namespace std;#define INFINITE 10000.00bool equal(double a, double b) {  //两个double类型数据不能简单地以==来判断相等return fabs(a - b) < 1e-6;}struct line {double k, b;double x1, y1, x2, y2;line() {}line(double xx1, double yy1, double xx2, double yy2, double kk, double bb): x1(xx1), y1(yy1), x2(xx2), y2(yy2), k(kk), b(bb) {}//  对线段(x1,y1)-(x2,y2)转化成 y=kx+b 形式，进行排序，按照 k 升序排序，如果 k 值相等，则按 b 值升序排序，//  如果 k,b都相等，则按 x1 升序排序，若 x1 还相等的话，则按 x2 升序排序。//  要对 平行于 y轴，即斜率不存在的线段 加以判断，//  先按照 b 值（b==x1）升序排序，再按 y1 升序排序，最后按 y2 升序排序bool operator<(const line& other) const{ //  自定义比较函数if (equal(k, other.k)){if (equal(k, INFINITE)) {if (equal(b, other.b)) {if (equal(y1, other.y1))return y2 < other.y2;elsereturn y1 < other.y1;}elsereturn b < other.b;}else if (equal(b, other.b)) {if (x1 == other.x1)return x2 < other.x2;elsereturn x1 < other.x1;}elsereturn b < other.b;}elsereturn k<other.k;}} s[10000];int main() {int lineNum;double x1, y1, x2, y2;while (cin >> lineNum && lineNum != 0) {for (int i = 0; i < lineNum; i++) {cin >> x1 >> y1 >> x2 >> y2;if (x1 > x2 || (equal(x1, x2) && y1 > y2)) {  //保证x1<x2或斜率不存在时有y1<y2swap(x1, x2); swap(y1, y2); }if (equal(x1, x2))s[i] = line(x1, y1, x2, y2,INFINITE,x1);else {s[i] = line(x1, y1, x2, y2, (y2 - y1) / (x2 - x1), 0);s[i].b = y1 - s[i].k * x1;}}sort(s, s + lineNum);int counter = 1;double farX = s[0].x2, farY = s[0].y2;  //（farX,farY）记录当前线段的终点for (int i = 1; i < lineNum; i++) {//如果k 或者b 不相等，则肯定不在同一直线上if (!equal(s[i].k, s[i - 1].k) || !equal(s[i].b, s[i - 1].b)) {farX = s[i].x2;farY = s[i].y2;counter++;}//在同一直线上但没有公共部分else if ((s[i].k != INFINITE && s[i].x1 > farX) || (s[i].k == INFINITE && s[i].y1 > farY)) {farX = s[i].x2;farY = s[i].y2;counter++;}//在同一直线上而且有公共部分，如果当前线段的终点更远，那也要更新(x,y)else if ((s[i].k != INFINITE && farX < s[i].x2) || (s[i].k == INFINITE && farY < s[i].y2)) {farX = s[i].x2;farY = s[i].y2;}}cout << counter << endl;}return 0;}