Segment set（并查集 + 计算几何问题）

最新推荐文章于 2019-04-21 15:15:46 发布

原创最新推荐文章于 2019-04-21 15:15:46 发布 · 392 阅读

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本文探讨了一种结合计算几何与并查集的数据结构问题，通过快速排斥实验和跨立实验来判断线段是否相交，并利用并查集维护线段集合的状态，最终求出特定线段集合的大小。

Segment set
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 253 Accepted Submission(s): 120

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
/*
这道题直接没有思路所以直接百度了别人AC的代码发现是计算几何问题
先给大家补充一下知识吧快速排斥实验和跨立实验
http://blog.sina.com.cn/s/blog_71dbfe2e0101f7zb.html
这里面介绍挺全面的我也从里面学会了跨立实验和快速排斥实验
快速排斥实验
关键代码：
min(a.x1,a.x2)<=max(b.x1,b.x2)&&
min(b.x1,b.x2)<=max(a.x1,a.x2)&&
min(a.y1,a.y2)<=max(b.y1,b.y2)&&
min(b.y1,b.y2)<=max(a.y1,a.y2)
通过了快速排斥实验之后还有一些特殊情况人需要判断所以这里就需要通过跨立实验
跨立实验关键代码
（(P3-P1)叉乘(P2-P1) ）点乘 ((P4 - P1) 叉乘 (P2 - P1)) >=0才可以由于只有当线段P3P4和P1P2跨立成功才可以认为相交
*/

#include<iostream>
#include<stdio.h>
#include<algorithm>
#include<string.h>

using namespace std;

typedef struct bian
{
    double x1,x2;
    double y1,y2;
    int father;
}node;

int rr[1007];
node t[1007];
int father(int a)
{
    while(a!=t[a].father)
        a = t[a].father;
    return a;
}
void Union_set(int a,int b)
{
    int pa = father(a);
    int pb = father(b);
    if(pa!=pb)
    {
        t[pa].father = pb;
        rr[pb] += rr[pa];
    }
}
int xiangjiao(node a,node b)
{
    if(min(a.x1,a.x2)<=max(b.x1,b.x2)&&min(b.x1,b.x2)<=max(a.x1,a.x2)&&min(a.y1,a.y2)<=max(b.y1,b.y2)&&min(b.y1,b.y2)<=max(a.y1,a.y2))
    {
        //跨立实验
        double t1 = (a.x2-a.x1)*(b.y1 - a.y1) - (b.x1 - a.x1 )*(a.y2 - a.y1);
        double t2 = (a.x2-a.x1)*(b.y2 - a.y1) - (b.x2 - a.x1 )*(a.y2 - a.y1);
        double t3 = (b.x2 - b.x1)*(a.y1 - b.y1) - (a.x1 - b.x1) *(b.y2 - b.y1);
        double t4 = (b.x2 - b.x1)*(a.y2 - b.y1) - (a.x2 - b.x1) *(b.y2 - b.y1);
        if(t1*t2<=0&&t3*t4<=0)
            return 1;
    }
    return 0;
}
int main()
{
    int ncase;
    cin>>ncase;
    while(ncase--)
    {
        int cmd;
        cin>>cmd;
        for(int i =1;i<=cmd;i++)
        {
            t[i].father = i;
            rr[i] = 1;
        }
        int sum = 1;
        while(cmd--)
        {
            string str;
            cin>>str;
            if(str[0]=='P')
            {
                cin>>t[sum].x1>>t[sum].y1>>t[sum].x2>>t[sum].y2;
                for(int i =1;i<sum;i++)
                {
                    if(xiangjiao(t[i],t[sum]))
                    {
                        Union_set(i,sum);
                    }
                }
                sum++;
            }else
            {
                int mm;
                cin>>mm;
                cout<<rr[father(mm)]<<endl;
            }
        }
        if(ncase)
            cout<<endl;

    }
    return 0;
}