hdu 1558 Segment set（判线段相交模版）

最新推荐文章于 2019-08-21 23:04:00 发布

KinFungYu

最新推荐文章于 2019-08-21 23:04:00 发布

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分类专栏：并查集文章标签：线段相交

本文链接：https://blog.youkuaiyun.com/WEYuLi/article/details/8979204

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解决线段集查询问题，通过并查集算法判断线段是否相连，并查询特定线段所在集合大小。输入包括多组测试数据，每组数据包含多个命令，命令包括绘制线段和查询线段集大小。

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Segment set

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2644 Accepted Submission(s): 994

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

并查集水题，这题要判线段相交。

AC代码：
#include <iostream>
#include <cstring>
#include <string>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <map>
#include <cmath>
#include <vector>
#include <cstdlib>
#define eps 1e-8
using namespace std;
const int MAX=1005;
int father[MAX],rank[MAX];
struct node
{
    double x1,y1;
    double x2,y2;
} line[1010];
int max(int a,int b)
{
    return a>b?a:b;
}
int min(int a,int b)
{
    return a<b?a:b;
}
double multiply1(node a,node b)
{
    return (a.x1-a.x2)*(b.y1-a.y1)-(a.y1-a.y2)*(b.x1-a.x1);
}
double multiply2(node a,node b)
{
    return (a.x1-a.x2)*(b.y2-a.y1)-(a.y1-a.y2)*(b.x2-a.x1);
}
bool inter(node a,node b)
{
    if(max(a.x1,a.x2)>=min(b.x1,b.x2)&&
            max(b.x1,b.x2)>=min(a.x1,a.x2)&&
            max(a.y1,a.y2)>=min(b.y1,b.y2)&&
            max(b.y1,b.y2)>=min(a.y1,a.y2)&&
            multiply1(a,b)*multiply2(a,b)<=eps&&
            multiply1(b,a)*multiply2(b,a)<=eps)
        return true;
    else return false;
}
int Find_set(int n)
{
    if(n!=father[n])
        father[n]=Find_set(father[n]);
    return father[n];
}
void Union(int a,int b)
{
    a=Find_set(a);
    b=Find_set(b);
    if(a==b) return;
    father[b]=a;
    rank[a]+=rank[b];
}
void init()
{
    for(int i=0; i<MAX; i++)
    {
        father[i]=i;
        rank[i]=1;
    }
}
int main()
{
    int t,n,m;
    char s[2];
    scanf("%d",&t);
    while(t--)
    {
        init();
        int cnt=1;
        scanf("%d",&n);
        while(n--)
        {
            scanf("%s",s);
            if(s[0]=='P')
            {
                scanf("%lf%lf%lf%lf",&line[cnt].x1,&line[cnt].y1,&line[cnt].x2,&line[cnt].y2);
                for(int i=1; i<cnt; i++)
                    if(inter(line[i],line[cnt]))
                        Union(i,cnt);
                cnt++;
            }
            else
            {
                scanf("%d",&m);
                int root=Find_set(m);
                printf("%d\n",rank[root]);
            }
        }
        if(t>0)
        printf("\n");
    }
    return 0;
}