bzoj 1007: [HNOI2008]水平可见直线

1007: [HNOI2008]水平可见直线

Time Limit: 1 Sec  Memory Limit: 162 MB
Submit: 7378  Solved: 2829
[Submit][Status][Discuss]

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

单调栈，所有直线按斜率从小到大排序，如果当前直线与栈顶直线的交点在栈顶两直线交点左边，

则栈顶直线一定不可见，弹出栈顶

http://www.lydsy.com/JudgeOnline/problem.php?id=1007

#include<stdio.h>#include<math.h>#include<string.h>#include<algorithm>using namespace std;typedef struct Res{double k;double b;int id;bool operator < (const Res &c) const{if(k<c.k || k==c.k && b<c.b)return 1;return 0;}}Res;Res s[50005];int st[50005], ans[50005];int main(void){int n, i, top;double nowx, dx;while(scanf("%d", &n)!=EOF){for(i=1;i<=n;i++){scanf("%lf%lf", &s[i].k, &s[i].b);s[i].id = i;}sort(s+1, s+n+1);top = 0;for(i=1;i<=n;i++){if(i<=n-1 && s[i].k==s[i+1].k)continue;while(top>=2){nowx = (s[st[top]].b-s[i].b)/(s[i].k-s[st[top]].k);dx = (s[st[top-1]].b-s[st[top]].b)/(s[st[top]].k-s[st[top-1]].k);if(nowx<=dx)top--;elsebreak;}st[++top] = i;}memset(ans, 0, sizeof(ans));for(i=1;i<=top;i++)ans[s[st[i]].id] = 1;for(i=1;i<=n;i++){if(ans[i]==1)printf("%d ", i);}printf("\n");}return 0;}