hdu 1558 Segment set【并查集+计算几何】
Segment set
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4460 Accepted Submission(s): 1665
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.
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
Author
LL
思路：并查集+判断集合个数+计算几何、思路来源：http://acm.hrbust.edu.cn/index.php?m=ProblemSet&a=showProblem&problem_id=1507【对应题解：
http://blog.csdn.net/mengxiang000000/article/details/51174476 】
思路：对于相交的两条直线，加入到一个集合里边，并对其元素个数进行维护，假设我们让x的祖先归到y下边：f【find（x）】=find（y）；同时我们用这样的操作来维护一个集合里边的元素个数：sum【find（y）】+=sum【find（x）】这样我们就能把一个集合里边一共有多少个元素归到祖先标号的数组里，所以如果查询当前节点有几个节点和他属于同一个集合的时候直接输出sum【find（当前节点）】即可、
注意的点：输出格式有点坑爹，各位童鞋注意一下。
AC代码：
#include<stdio.h>
#include<iostream>
#include<string.h>
using namespace std;
struct point
{
double x,y;
}b[5];
struct xianduan
{
double x1,x2,y1,y2;
}a[10000];
int f[10000];
int sum[10000];
int find(int x)
{
return f[x] == x ? x : (f[x] = find(f[x]));
}
void merge(int a,int b)
{
int A,B;
A=find(a);
B=find(b);
if(A!=B)
{
f[B]=A;
sum[A]+=sum[B];
}
}
double multi(point p1,point p2,point p0)
{
return((p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y));
}
bool IsIntersected(point s1,point e1,point s2,point e2)//两个线段相交
{
return(max(s1.x,e1.x)>=min(s2.x,e2.x))&&
(max(s2.x,e2.x)>=min(s1.x,e1.x))&&
(max(s1.y,e1.y)>=min(s2.y,e2.y))&&
(max(s2.y,e2.y)>=min(s1.y,e1.y))&&
(multi(s1,s2,e1)*multi(s1,e1,e2)>=0)&&
(multi(s2,s1,e2)*multi(s2,e2,e1)>=0);
}
int main()
{
int t;
scanf("%d",&t);
int kase=0;
while(t--)
{
if(kase)
printf("\n");
for(int i=0;i<10000;i++)
{
f[i]=i;sum[i]=1;
}
int q;
scanf("%d",&q);
int cont=0;
while(q--)
{
char s[5];
scanf("%s ",s);
if(s[0]=='P')
{
scanf("%lf%lf%lf%lf",&a[cont].x1,&a[cont].y1,&a[cont].x2,&a[cont].y2);
b[0].x=a[cont].x1;b[0].y=a[cont].y1;b[1].x=a[cont].x2;b[1].y=a[cont].y2;
for(int i=0;i<cont;i++)
{
b[2].x=a[i].x1;b[2].y=a[i].y1;
b[3].x=a[i].x2;b[3].y=a[i].y2;
if(IsIntersected(b[0],b[1],b[2],b[3]))
{
merge(i,cont);
}
}
cont++;
}
if(s[0]=='Q')
{
int k;
scanf("%d",&k);
k--;
printf("%d\n",sum[find(k)]);
}
}
kase++;
}
}