【BZOJ1007】[HNOI2008]水平可见直线

【BZOJ1007】[HNOI2008]水平可见直线

题面

bzoj

洛谷

题解

这题是用来搞笑的吧。。。

一眼看来就是半平面交啊，结果发现是个弱化版半平面交。

就按照半平面交的方法就行了。

代码

#include <iostream> 
#include <cstdio> 
#include <cstdlib> 
#include <cstring> 
#include <cmath> 
#include <algorithm> 
using namespace std; 
inline int gi() { 
    register int data = 0, w = 1; 
    register char ch = 0; 
    while (!isdigit(ch) && ch != '-') ch = getchar(); 
    if (ch == '-') w = -1, ch = getchar(); 
    while (isdigit(ch)) data = 10 * data + ch - '0', ch = getchar(); 
    return w * data; 
} 
const int MAX_N = 5e4 + 5; 
struct Line { int k, b, id; } ; 
bool operator < (const Line &l, const Line &r)
    { return l.k == r.k ? l.b > r.b : l.k < r.k; }
double Intersection(Line &l, Line &r) 
    { return 1.0 * (l.b - r.b) / (r.k - l.k); } 
bool cmp_i(const Line &l, const Line &r) { return l.id < r.id; } 
int N; 
Line l[MAX_N], stk[MAX_N]; 

int main () {
#ifndef ONLINE_JUDGE 
    freopen("cpp.in", "r", stdin); 
#endif 
	N = gi(); for (int i = 1; i <= N; i++) l[i] = (Line){gi(), gi(), i}; 
	sort(&l[1], &l[N + 1]); 
	l[0].k = -1e9; 
	int cnt = 0;
	for (int i = 1; i <= N; i++) if (l[i].k != l[i - 1].k) l[++cnt] = l[i]; 
	N = cnt;
	int top = 0; 
	stk[++top] = l[1]; 
	for (int i = 2; i <= N; i++) { 
		while (top > 1 && Intersection(l[i], stk[top]) <= Intersection(l[i], stk[top - 1])) --top; 
		stk[++top] = l[i]; 
	} 
	sort(&stk[1], &stk[top + 1], cmp_i);  
	for (int i = 1; i <= top; i++)  printf("%d ", stk[i].id); 
	printf("
"); 
	return 0; 
}