zoj 2967 Colorful Rainbows求不被覆盖的直线数(贪心)

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

2

1
1 1

3
1 0
2 0
3 0

Sample Output

1
2
http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=1966
#include<iostream>
#include<algorithm>
#include<string>
#include<map>
#include<string.h>
#include<vector>
#include<cmath>
#include<stdlib.h>
#include<cstdio>
#define ll long long
using namespace std;
int comp(pair<double,double>a,pair<double,double>b){
	if(a.first!=b.first)
		return a.first<b.first;
	return a.second<b.second;
}
int main(){
	int t;
	cin>>t;
	while(t--){
		int n;
		cin>>n;
		pair<double,double> xian[6000];
		for(int i=0;i<n;i++){
			double a,b;
			cin>>a>>b;
			xian[i]=make_pair(a,b); //存每条线的斜率和 b 
		}
		sort(xian,xian+n,comp);
		pair<double,int>jiao;
		jiao=make_pair(10000007,n-1);  //从最后一个点开始 
		int sum=1;
		while(1){
			double maxn = -100000010;
			int biao=-1;
			int u=0;
			for(int i=0;i<n-1;i++){
				if(xian[i].first ==1.1&&xian[i].second ==1.1)
					continue;
				double x = (xian[i].second-xian[jiao.second].second)/(xian[jiao.second].first-xian[i].first);
			 	if(x<jiao.first&&x>maxn){ //x是两条线段交点的横坐标 
			 		maxn = x;
			 		biao = i;
			 		u=1;
			 	}
				else if(x==jiao.first&&xian[i].first<xian[jiao.second].first){  //x==jiao.first当前交点与上一次的交点横坐标一样时，去斜率较小的 
			 		maxn = x;
			 		biao = i;
			 		u=1;
			 	}
			}
			if(u==0)
				break;
			xian[jiao.second]=make_pair(1.1,1.1);  //表示该条直线移除 
			jiao = make_pair(maxn,biao);  //最大的横坐标 ，下一次的目标直线下标 
			sum++;
		}	
		cout<<sum<<endl;	 
	}
	return 0;
}