Segment set
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5413 Accepted Submission(s): 2087
Problem Description
A segment and all segments which are connected with it compose a segment set. The size of a segment set is the number of segments in it. The problem is to find the size of some segment set.
Input
In the first line there is an integer t - the number of test case. For each test case in first line there is an integer n (n<=1000) - the number of commands.
There are two different commands described in different format shown below:
P x1 y1 x2 y2 - paint a segment whose coordinates of the two endpoints are (x1,y1),(x2,y2).
Q k - query the size of the segment set which contains the k-th segment.
k is between 1 and the number of segments in the moment. There is no segment in the plane at first, so the first command is always a P-command.
Output
For each Q-command, output the answer. There is a blank line between test cases.
Sample Input
1
10
P 1.00 1.00 4.00 2.00
P 1.00 -2.00 8.00 4.00
Q 1
P 2.00 3.00 3.00 1.00
Q 1
Q 3
P 1.00 4.00 8.00 2.00
Q 2
P 3.00 3.00 6.00 -2.00
Q 5
Sample Output
1
2
2
2
5
题意比较好理解，增加一条线段，若果和已有的线段相交，则把他们加入到一个集合中，对于每个问讯，输出这条线段所在集合的元素个数；
思路：快速排斥跨立实验判断相交，计数并查集维护集合合并；
排斥实验的要点：
bool quick(double x1,double y1,double x2,double y2,double x3,double y3,double x4,double y4)
{
if(min(x1,x2)>max(x3,x4)) return false;
if(max(x1,x2)<min(x3,x4)) return false;
if(min(y1,y2)>max(y3,y4)) return false;
if(max(y1,y2)<min(y3,y4)) return false;
return true;
}排斥实验和跨立实验缺一不可
跨立实验要点：1，差集相乘的符号；2，必须要判断两次（两条线段交换判断）
ac代码：
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
int vis[100010];
int t;
int n;
int tot;
struct site
{
double x,y;
};
struct segment
{
site l,r;
int no;
}a[1005];
int pre[1005];
int cnt[1005];
int find(int x)
{
int rt=x;
while(rt!=pre[rt])
{
rt=pre[rt];
}
int i=x;
while(i!=pre[i])
{
int fa=pre[i];
pre[i]=rt;
i=fa;
}
return rt;
}
bool quick(double x1,double y1,double x2,double y2,double x3,double y3,double x4,double y4)
{
if(min(x1,x2)>max(x3,x4)) return false;
if(max(x1,x2)<min(x3,x4)) return false;
if(min(y1,y2)>max(y3,y4)) return false;
if(max(y1,y2)<min(y3,y4)) return false;
return true;
}
double cha(double x1,double y1,double x2,double y2)
{
return x1*y2-x2*y1;
}
double chaji(site a,segment b)
{
/* cout<<a.x<<' '<<a.y<<endl;
cout<<b.l.x<<' '<<b.l.y<<endl;
cout<<b.r.x<<' '<<b.r.y<<endl;
cout<<a.x-b.l.x<<' ';
cout<<a.y-b.l.y<<endl;
cout<<b.r.x-b.l.x<<' ';
cout<<b.r.y-b.l.y<<endl;*/
double t= cha(a.x-b.l.x, a.y-b.l.y , b.r.x-b.l.x , b.r.y-b.l.y);
// cout<<t<<endl<<endl;
return t;
}
bool judge(segment a,segment b)
{
//快速排斥实验
if(quick(a.l.x,a.l.y,a.r.x,a.r.y,b.l.x,b.l.y,b.r.x,b.r.y)==false) return false;
//跨立实验
if(chaji(a.l,b)*chaji(a.r,b)<=0&&chaji(b.l,a)*chaji(b.r,a)<=0) return true;
return false;
}
int main()
{
char s[4];
scanf("%d",&t);
while(t--)
{
tot=0;
scanf("%d",&n);
memset(cnt,0,sizeof cnt);
for(int i=0;i<n;i++)
{
pre[i]=i;
cnt[i]=1;
}
while(n--)
{
scanf("%s",s);
if(s[0]=='P')
{
scanf("%lf%lf%lf%lf",&a[tot].l.x,&a[tot].l.y,&a[tot].r.x,&a[tot].r.y);
for(int i=0;i<tot;i++)
{
if(judge(a[i],a[tot]))
{
int ri=find(i);
int rt=find(tot);
if(ri!=rt)
{
pre[rt]=ri;
cnt[ri]+=cnt[rt];
//cout<<i<<'#'<<endl;
}
}
}
tot++;
}
else if(s[0]=='Q')
{
int temp;
scanf("%d",&temp);
cout<<cnt[find(temp-1)]<<endl;
}
}
if(t) cout<<endl;
}
return 0;
}