HDU 1158 Segment set（线段相交并查集）_并查集解决线段交替问题-优快云博客

本文介绍了一种通过并查集实现的线段集查询算法，包括如何判断线段相交及更新线段集大小的方法。文章详细展示了算法的实现过程，并提供了一段完整的示例代码。

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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

这道题判断线段相交的部分不太容易，感觉判断线段相交才是主要知识点的样子。。。

判断了线段相交，其实数据量也不大，两层for循环可解决。最后就是并查集，其实多开一个数组记录当前线段所属集合的线段数量就可以了。

刚开始忘了在更新线段数量之后更新整个集合的线段数量，检查花了点时间。

代码：

#include<stdio.h>
#include<algorithm>
using namespace std;
int fa[1005],val[1005];
struct Point
{
    double x,y;
};
struct node
{
    Point a,b;
}po[1005];
int findd(int x)
{
    if(x==fa[x])
        return x;
    fa[x]=findd(fa[x]);
    val[x]=val[fa[x]];
    return fa[x];
}

double multiply(Point p1,Point p2,Point p0)
{
    return((p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y));
}

bool intersect(node u,node v)
{
    return( (max(u.a.x,u.b.x)>=min(v.a.x,v.b.x))&&
    (max(v.a.x,v.b.x)>=min(u.a.x,u.b.x))&&
    (max(u.a.y,u.b.y)>=min(v.a.y,v.b.y))&&
    (max(v.a.y,v.b.y)>=min(u.a.y,u.b.y))&&
    (multiply(v.a,u.b,u.a)*multiply(u.b,v.b,u.a)>=0)&&
    (multiply(u.a,v.b,v.a)*multiply(v.b,u.b,v.a)>=0));
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n,num=0;
        scanf("%d",&n);
        for(int i=0;i<=n;i++)
        {
            fa[i]=i;
            val[i]=1;
        }
        for(int i=0;i<n;i++)
        {
            char s[2];
            scanf("%s",s);
            if(s[0]=='P')
            {
                node mid;
                scanf("%lf%lf%lf%lf",&mid.a.x,&mid.a.y,&mid.b.x,&mid.b.y);
                for(int j=0;j<num;j++)
                {
                    if(intersect(mid,po[j]))
                    {
                        int id=findd(j);
                        if(fa[num]==num)
                        {
                            fa[num]=id;
                            val[id]+=val[num];
                        }
                        else
                        {
                            if(id!=fa[num])
                            {
                                fa[fa[num]]=id;
                                val[id]+=val[fa[num]];
                            }
                        }
                        findd(num);
                    }
                }
                po[num++]=mid;
            }
            else
            {
                int id;
                scanf("%d",&id);
                id=findd(id-1);
                printf("%d\n",val[id]);
            }
        }
        if(t)
            printf("\n");
    }
    return 0;
}