hdu 1558 Segment set(线段相交+并查集)

题目：​​http://acm.hdu.edu.cn/showproblem.php?pid=1558​​

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

题目叙述很清楚，相交的线段属于同一集合，问一共有多少个集合。判断线段相交是一个点，集合数目问题又是一个点。把这两个点搞定就能解决问题。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

int pre[1010],sum[1010];
struct point{
    double x,y;
};
struct node{
    point a,b;
} edge[1010];
int myedges;

int Find(int x){
    if(x!=pre[x]){
        pre[x]=Find(pre[x]);
    }
    return pre[x];
}

void Merge(int a,int b){
    int x=Find(a),y=Find(b);
    if(x!=y){
        pre[y]=x;
        sum[x]+=sum[y];
    }
}

double crossmulti(point a,point b,point c){
    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共线时使用
    if(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))return 1;
    return 0;
}

bool Cross(point a,point b,point c,point d){ //判断ab 与cd是否相交
    double re1,re2,re3,re4;
    re1=crossmulti(c,d,a);
    re2=crossmulti(c,d,b);
    re3=crossmulti(a,b,c);
    re4=crossmulti(a,b,d);
    if(re1*re2<0&&re3*re4<0)    return 1;
    else    if(re1==0&&OnSegment(c,d,a))    return 1;  //四种在同一条线上的结果
    else    if(re2==0&&OnSegment(c,d,b))    return 1;
    else    if(re3==0&&OnSegment(a,b,c))    return 1;
    else    if(re4==0&&OnSegment(a,b,d))    return 1;
    return 0;
}

int main()
{
    //freopen("cin.txt",&qumt;r",stdin);
    int i,j,k,t,n;
    cin>>t;
    while(t--)
    {
        int i,j,k,n;
        char s[10];
        scanf("%d",&n);
        myedges=0;
        for(i=1;i<=n;i++){
            pre[i]=i;
            sum[i]=1;
        }
        for(i=1;i<=n;i++)
        {
            scanf("%s",s);
            if(s[0]=='P'){
                myedges++;
                scanf("%lf%lf%lf%lf",&edge[myedges].a.x,&edge[myedges].a.y,&edge[myedges].b.x,&edge[myedges].b.y);
                for(j=1;j<myedges;j++)
                    if(Find(myedges)!=Find(j)&&Cross(edge[myedges].a,edge[myedges].b,edge[j].a,edge[j].b))
                        Merge(myedges,j);
            }
            else if(s[0]=='Q'){
                scanf("%d",&k);
                printf("%d\n",sum[Find(k)]);
            }
        }
        if(t>0) printf("\n");
    }
    return 0;
}