ZOJ2967-Colorful Rainbows

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本文介绍了一个有趣的几何问题——如何确定在无限大的纸上绘制的多条彩色彩虹中，哪些彩虹能够被看到。通过数学公式定义每条彩虹，并使用编程实现了解决方案，包括去重、排序及栈操作来找出可见的彩虹。

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Colorful Rainbows

Time Limit: 2 Seconds Memory Limit: 65536 KB

Evelyn likes drawing very much. Today, she draws lots of rainbows on white paper of infinite size, each using a different color. Since there're too many rainbows now, she wonders, how many of them can be seen?

For simplicity, each rainbow L_i is represented as a non-vertical line specified by the equation: y=a_ix+b_i. A rainbow L_i can be seen if there exists some x-coordinate x₀ at which, its y-coordinate is strictly greater than y-coordinates of any other rainbows: a_ix₀+b_i > a_jx₀+b_j for all j != i.

Now, your task is, given the set of rainbows drawn, figure out the number of rainbows that can be seen.

Input

Standard input will contain multiple test cases. The first line of the input is a single integer T (1 <= T <= 60) which is the number of test cases. And it will be followed by T consecutive test cases.

There's a blank line before every case. In each test case, there will first be an integer n (1 <= n <= 5000), which is the number of rainbows. Then n consecutive real number pairs follow. Each pair contains two real numbers, a_i and b_i, representing rainbow L_i: y=a_ix+b_i. No two rainbows will be the same, that is to say, have the same a and b.

Output

Results should be directed to standard output. The output of each test case should be a single integer, which is the number of rainbows that can be seen.

Sample Input

Sample Output

1
2

Author: DAI, Wenbin
Source: The 5th Zhejiang Provincial Collegiate Programming Contest

题意：给出n条y=ai*x+bi的直线。对于这些直线，如果存在x使得该直线y大于其他任意一直线，那么这条直线可以被看见，问有多少条直线可以被看见。

解题思路：首先去重，将那些a值相同的直线取其中b最大的那条保留下来，其他的全删掉。再将直线按照a值从小到大排序，因为斜率不同，所以任意两条直线都会相交。而这些直线是按照斜率从小到大进行排序，所以当x小于其交点x值时，斜率小的y值大。利用这一特性将，先让线入栈。若将入栈的线与栈顶线的交点x值小于栈顶两条线的的交点的x值，则将栈顶线出栈，继续进行上一次判断，直到所有线都判定过，第一条入栈的直线交点可以判断为-INF。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <cmath>
#include <algorithm>
#include <stack>
#include <queue>
#include <climits>
#include <functional>
#include <vector>
#include <map>
#include <set>

using namespace std;

#define LL long long
const int INF=0x3f3f3f3f;

struct node
{
    double a;
    double b;
} s[5010],x[5010];

bool cmp(node a1,node a2)
{
    if(a1.a!=a2.a) return a1.a<a2.a;
    return a1.b<a2.b;
}

int main()
{
    int t,n,ans;
    double d,p;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(int i=0; i<n; i++) scanf("%lf%lf",&s[i].a,&s[i].b);
        sort(s,s+n,cmp);
        int cnt=0;
        for(int i=0; i<n-1; i++)
        {
            if(s[i].a==s[i+1].a) continue;
            s[cnt++]=s[i];
        }
        s[cnt++]=s[n-1];
        x[0]=s[0];
        d=-INF;
        ans=1;
        for(int i=1; i<cnt; i++)
        {
            p=(s[i].b-x[ans-1].b)*1.0/(x[ans-1].a-s[i].a);
            while(p<=d)
            {
                ans--;
                if(ans>1)
                {
                    d=(x[ans-1].b-x[ans-2].b)*1.0/(x[ans-2].a-x[ans-1].a);
                    p=(s[i].b-x[ans-1].b)*1.0/(x[ans-1].a-s[i].a);
                }
                else break;
            }
            d=(s[i].b-x[ans-1].b)*1.0/(x[ans-1].a-s[i].a);
            x[ans++]=s[i];
        }
        printf("%d\n",ans);
    }
    return 0;
}