2017 Wuhan University Programming Contest 现场赛G. Room（set+思路）

问题 G: Room

时间限制: 1 Sec  内存限制: 64 MB
提交: 11  解决: 8
[提交][状态][讨论版]

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

30 200000 1000000 600000600000 0 300000 10000000 800000 1000000 4000009200000 900000200000 400000300000 600000600000 500000700000 700000800000 300000300000 200000800000 100000600000 200000

样例输出

6

提示

[提交][状态]题意：给出n个起始点和终止点，保证每个点都在边上，每两个点连成一条线，把矩形分成若干个区域，现在有t个点有人，在有人的区域内建设服务点，一个区域只建一个就够了，问最少建多少个，也就是问有多少个有人的区域思路：判断每个点与所有直线的关系，分上下两种关系，分别用1，0标记，把每个点与所有直线的关系都存到字母串里，然后插入到set，这样重复的排列，也就是相同区域的点不重复插入，也就是最小，判断与直线的关系可以用两点式求出方程，然后将点带入判断正负

#include <iostream>#include <cstdio>#include <cstring>#include<string.h>#include <algorithm>#include<set>#define ll long long#define wtf printf("wtf");using namespace std;struct node{    int x1,y1,x2,y2;}e[100005];struct xx{    int x,y;}p[105];set<string>s;ll check(node a,xx b){    ll A=a.y2-a.y1;    ll B=a.x1-a.x2;    ll C=(-B*a.y1-A*a.x1);    return A*b.x+B*b.y+C>0;}ll solve(ll n,ll t){    for(ll i=1;i<=t;i++)    {             string str="";        for(ll j=1;j<=n;j++)        str+=check(e[j],p[i])+'0';         s.insert(str);    }    return s.size();}int main(){    ll n,t;    cin>>n;    for(ll i=1;i<=n;i++)        scanf("%lld%lld%lld%lld",&e[i].x1,&e[i].y1,&e[i].x2,&e[i].y2);    cin>>t;    for(ll i=1;i<=t;i++)    scanf("%lld%lld",&p[i].x,&p[i].y);   printf("%lld\n",solve(n,t));   return 0;}