1007: [HNOI2008]水平可见直线 Time Limit:1 Sec Memory Limit:162 MB Submit:1830 Solved:656 [ Submit ][ Status ][ Discuss ] Description Input 第一行为N(0 N 50000),接下来的N行输入Ai,Bi Out
1007: [HNOI2008]水平可见直线
Time Limit: 1 Sec
Memory Limit: 162 MB
Submit: 1830
Solved: 656
[Submit][Status][Discuss]
Description
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，从小到大插入。
半平面交的特殊情况：
每次
都要保证x坐标<x,top>><top,top'> 否则top不可见（top为栈顶元素）
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<functional>
#include<iostream>
using namespace std;
#define MAXN (50000+10)
int n;
struct line
{
int k,b,i;
friend bool operator<(line a,line b) {return (a.k==b.k)?a.b>b.b:a.k<b.k; }
friend double intx(line a,line b)
{
return (double)(b.b-a.b)/(a.k-b.k);
}
}a[MAXN];
int s[MAXN],size=0;
void push(int x)
{
while (size>1&&intx(a[s[size]],a[s[size-1]])>=intx(a[s[size]],a[x])) size--;
s[++size]=x;
}
bool b[MAXN];
int main()
{
scanf("%d",&n);
for (int i=1;i<=n;i++) {scanf("%d%d",&a[i].k,&a[i].b);a[i].i=i;}
sort(a+1,a+1+n);
push(1);
for (int i=2;i<=n;i++)
if (a[i].k>a[i-1].k) push(i);
// for (int i=1;)
memset(b,0,sizeof(b));for (int i=1;i<=size;i++) b[a[s[i]].i]=1;
for (int i=1;i<=n;i++) if (b[i]) cout<<i<<' ';
return 0;
}