room [判断点与线的关系]

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时间限制: 1 Sec   内存限制: 64 MB
提交: 16   解决: 13
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题目描述

The ACM / ICPC team has a large room, the length and width of which is 10^6 . However, the guys in ACM / ICPC teams are too lazy to make their study room tidy. So there are wires everywhere and divide the room into several parts. A team in a part of the room cannot move out of it or they might touch the wires and the network will down. To make every team can compete in the contest, they have to set up some facilities such as toilet since the teams should do anything in their parts. Now, we will give you the map of our study room and the position of the teams, your task is to calculate how many facilities is required to let every team can access a facilities to finish the contest without move out of their part. You should note that two or more teams can share a facility. 

输入

输出

Output one integer, the minimal facilities required.

样例输入

3
0 200000 1000000 600000
600000 0 300000 1000000
0 800000 1000000 400000
9
200000 900000
200000 400000
300000 600000
600000 500000
700000 700000
800000 300000
300000 200000
800000 100000
600000 200000

样例输出

6

提示

题意：

给你n条线，把整个地图分成若干份，

然后给你m个队伍。 问你这些队伍最多占领了地图的多少份，

题解：

一个点与线的关系只有三种，而题中保证不在线上，那只有线上方or 下方。

然后对线排个序，然后求出一个点到所有线的关系，存进string 然后用set统计即可。

、、

#include<string>
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<set>
#define ll long long
#define x first
#define y second
using namespace std;
const int mod=11;
const int maxn=100100;
const int maxm=111;
typedef pair<ll,ll>point;
typedef pair<point,point>li;
li line[maxn];
set<string>s;
int n,m;

int judge(li a,point p){
    ll A=a.second.y-a.first.y;
    ll B=a.first.x-a.second.x;
    ll C=A*a.first.x+B*a.first.y;
    return p.x*A+p.y*B-C>0;
}
int main(){
    scanf("%d",&n);
    point p1,p2,p3;
    for(int i=1;i<=n;++i){
        scanf("%lld %lld %lld %lld",&p1.x,&p1.y,&p2.x,&p2.y);
        line[i].x=p1,line[i].y=p2;
    }
    scanf("%d",&m);
    for(int i=1;i<=m;++i){
        scanf("%lld %lld",&p1.x,&p1.y);
        string str="";
        for(int j=1;j<=n;++j){
            str+=judge(line[j],p1)+'0';
        }
        s.insert(str);
    }
    printf("%d\n",s.size());
	return 0;
}