hdu 1558 Segment set(并查集+计算几何线段相交)

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用到的计算几何中的内容至今不大懂，目前练习并查集，以后待续、、、、

 

1、题目大意：

给你一些操作，P后边输入四个值，分别代表一条线段的起点、终点坐标，

当输入Q时，后边输入一个整形值K，输出第k条线段所在的集合中包含的线段的个数

2、思路：用并查集做

当输入P时，判断后边输入的线段的起点和终点时，判断跟之前的线段有没有相交，如果有相交，就merge（）合并，

如果输入的是Q时，就打印出当前线段所在集合的个数

3、题目：

Segment set
Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2386    Accepted Submission(s): 924


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
 

Author
LL
 

Source
HDU 2006-12 Programming Contest 
 

Recommend

3、代码：

#include<stdio.h>
#include<iostream>
using namespace std;
int set[1010];
int num[1010];
struct point
{
    double x,y;
};
struct edge
{
    point a;
    point b;
}e[1010];
int find(int x)
{
    int r=x;
    while(r!=set[r])
    r=set[r];
    int i=x;
    while(i!=r)
    {
       int j=set[i];
        set[i]=r;
        i=j;
    }
    return r;
}

void merge(int x,int y)
{
    int fx=find(x);
    int fy=find(y);
    if(fx!=fy)
    {
        set[fx]=fy;
        num[fy]+=num[fx];
    }
}
double xmult(point a,point b,point c) //大于零代表a,b,c左转
{
    return (b.x-a.x)*(c.y-a.y)-(b.y-a.y)*(c.x-a.x);
}
bool OnSegment(point a,point b,point c) 		//a,b,c共线时有效
{
    return c.x>=min(a.x,b.x)&&c.x<=max(a.x,b.x)&&c.y>=min(a.y,b.y)&&c.y<=max(a.y,b.y);
}

bool Cross(point a,point b,point c,point d) //判断ab 与cd是否相交
{
    double d1,d2,d3,d4;
    d1=xmult(c,d,a);
    d2=xmult(c,d,b);
    d3=xmult(a,b,c);
    d4=xmult(a,b,d);
    if(d1*d2<0&&d3*d4<0)	return 1;
    else	if(d1==0&&OnSegment(c,d,a))	return 1;
    else	if(d2==0&&OnSegment(c,d,b))	return 1;
    else	if(d3==0&&OnSegment(a,b,c))	return 1;
    else	if(d4==0&&OnSegment(a,b,d))	return 1;
    return 0;
}
int main()
{
    int t,n,k,tmp,i,j;
    char s[10];
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        k=0;
        for(int i=1;i<=n;i++)
        {
            set[i]=i;
            num[i]=1;
        }
        for(int i=1;i<=n;i++)
        {
            scanf("%s",s);
            if(s[0]=='P')
            {
                k++;
                scanf("%lf%lf%lf%lf",&e[k].a.x,&e[k].a.y,&e[k].b.x,&e[k].b.y);
                for(int j=1;j<k;j++)
                {
                    if(find(k)!=find(j)&&Cross(e[k].a,e[k].b,e[j].a,e[j].b))
                    merge(k,j);
                }
            }
            else if(s[0]=='Q')
            {
                scanf("%d",&tmp);
                printf("%d\n",num[find(tmp)]);
            }
        }
        if(t)//There is a blank line between test cases.注意最后一个样例不用输出空行
        printf("\n");
    }
    return 0;
}
/*
2
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
3
P 1.00 1.00 4.00 2.00
P 1.00 -2.00 8.00 4.00
Q 1
*/