BZOJ-1007 水平可见直线

简单几何题。。。用个栈维护当前可见直线，然后通过判断三直线的交点的左右即可进行选择。

#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <fstream>
#include <iostream>
#include <queue>
#define rep(i, l, r) for(int i=l; i<=r; i++)
#define N 56789
#define MAX 1<<30
#define clr(x, c) memset(x, c, sizeof(x))
#define ll long long
using namespace std;
int read()
{
	ll x=0, f=1; char ch=getchar();
	while (ch<'0' || ch>'9') { if (ch=='-') f=-1; ch=getchar(); }
	while (ch>='0' && ch<='9') { x=x*10+ch-'0'; ch=getchar(); }
	return x*f;
}
struct node{ll a, b, n;} c[N], v[N];
int n, vv;

bool cmp(node a, node b) { return a.a<b.a || (a.a==b.a && a.b>b.b); }
bool cmp2(node a, node b) { return a.n<b.n; }

int main()
{
	n=read();
	rep(i, 1, n) c[i].a=read(), c[i].b=read(), c[i].n=i; sort(c+1, c+1+n, cmp);
	v[vv=1]=c[1]; rep(i, 2, n) if (c[i].a != c[i-1].a)
	{
		while (vv>1 && (v[vv].a-c[i].a)*(v[vv].b-v[vv-1].b) >= (v[vv-1].a-v[vv].a)*(c[i].b-v[vv].b)) vv--;
		v[++vv]=c[i];
	}
	sort(v+1, v+1+vv, cmp2);
	rep(i, 1, vv) printf("%lld ", v[i].n);
	return 0;
}