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

110P 1.00 1.00 4.00 2.00P 1.00 -2.00 8.00 4.00Q 1P 2.00 3.00 3.00 1.00Q 1Q 3P 1.00 4.00 8.00 2.00Q 2P 3.00 3.00 6.00 -2.00Q 5

 

Sample Output

12225

 

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;elseif(re1==0&&OnSegment(c,d,a))return 1;  //四种在同一条线上的结果elseif(re2==0&&OnSegment(c,d,b))return 1;elseif(re3==0&&OnSegment(a,b,c))return 1;elseif(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;}