BZOJ1007 HNOI2008 水平可见直线

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1007: [HNOI2008]水平可见直线

Time Limit: 1 Sec Memory Limit: 162 MB
Submit: 4609 Solved: 1703
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Description

在xoy直角坐标平面上有n条直线L1,L2,...Ln,若在y值为正无穷大处往下看,能见到Li的某个子线段,则称Li为可见的,否则Li为被覆盖的.
例如,对于直线:
L1:y=x; L2:y=-x; L3:y=0
则L1和L2是可见的,L3是被覆盖的.
给出n条直线,表示成y=Ax+B的形式(|A|,|B|<=500000),且n条直线两两不重合.求出所有可见的直线.

Input

第一行为N(0 < N < 50000),接下来的N行输入Ai,Bi

Output

从小到大输出可见直线的编号，两两中间用空格隔开,最后一个数字后面也必须有个空格

Sample Input

3
-1 0
1 0
0 0

Sample Output

1 2

按斜率排序，然后维护一个下凸壳，记得去重。
代码如下：

/* * @Author: duyixian* @Date:   2015-09-08 17:00:18* @Last Modified by:   duyixian* @Last Modified time: 2015-09-10 11:10:38*/#include "cstdio"#include "cstdlib"#include "iostream"#include "algorithm"#include "cstring"#include "queue"#include "cmath"using namespace std;#define MAX_SIZE 50005#define INF 0x3F3F3F3F#define Eps 1e-5#define Modinline int Get_Int(){int Num = 0, Flag = 1;char ch;do{ch = getchar();if(ch == '-')Flag *= -1;}while(ch < '0' || ch > '9');do{Num = Num * 10 + ch - '0';ch = getchar();}while(ch >= '0' && ch <= '9');return Num * Flag;}struct Line{double a, b;int Num;inline bool operator < (Line const &temp) const{if(a < temp.a - Eps)return true;else if(a > temp.a + Eps)return false;elsereturn b > temp.b;}}Lines[MAX_SIZE], A[MAX_SIZE], Stack[MAX_SIZE];int N, Total, Top;bool Ans[MAX_SIZE];inline double Count(Line x, Line y){return (x.b - y.b) / (y.a - x.a);}int main(){cin >> N;for(int i = 1; i <= N; ++i){scanf("%lf%lf", &Lines[i].a, &Lines[i].b);Lines[i].Num = i;}sort(Lines + 1, Lines + N + 1);for(int i = 1; i <= N; ++i)if(abs(Lines[i].a - Lines[i - 1].a) > Eps)A[++Total] = Lines[i];Stack[++Top] = A[1];Stack[++Top] = A[2];for(int i = 3; i <= Total; ++i){while(Top >= 2 && Count(A[i], Stack[Top]) <= Count(Stack[Top], Stack[Top - 1]) + Eps)--Top;Stack[++Top] = A[i];}for(int i = 1; i <= Top; ++i)Ans[Stack[i].Num] = true;for(int i = 1; i <= N; ++i)if(Ans[i])printf("%d ", i);return 0;}

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