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hdu 1558 线段相交+并查集路径压缩

2024-07-19 01:05:15   223人阅读

Segment set

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3457    Accepted Submission(s): 1290


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
12225
 

 

Author
LL
 

 

Source
HDU 2006-12 Programming Contest
 
题目大意：有n个指令，p加入一条线段,q查询id线段所在集合（两线段有交点为同一集合）的元素个数。
思路：用并查集路径压缩记录各个线段间的关系，根据叉积的定义有：Cross(v,w)=0时w在v上,>0时w在v上方,<0时w在v下方。
两线段有交点的必要条件：必须每条线段的两个端点在另一线段的两侧或直线上。
 1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cmath> 5 using namespace std; 6  7 const double eps=1e-8; 8 const int maxn=1005; 9 int f[maxn];10 struct Point11 {12     double x,y;13     Point(){}14     Point(double x,double y):x(x),y(y){}15 };16 struct Line17 {18     Point a,b;19 }L[maxn];20 typedef Point Vector;21 Vector operator -(Vector A,Vector B){return Vector(A.x-B.x,A.y-B.y);}22 int dcmp(double x)23 {24     if(fabs(x)<eps) return 0;25     else return x<0?-1:1;26 }27 double Cross(Vector A,Vector B){ return A.x*B.y-A.y*B.x;}//叉积28 29 bool judge(Line a,Line b)//Cross(v,w)=0时w在v上,>0时w在v上方,<0时w在v下方30 {31     if(dcmp(Cross(a.a-b.a,b.b-b.a)*Cross(a.b-b.a,b.b-b.a))<=032         &&dcmp(Cross(b.a-a.a,a.b-a.a)*Cross(b.b-a.a,a.b-a.a))<=0)33         return true;34     return false;35 }36 int findset(int x){return f[x]!=x?f[x]=findset(f[x]):x;}37 void Union(int a,int b)38 {39     a=findset(a);b=findset(b);40     if(a!=b) f[a]=b;41 }42 int main()43 {44     int t,n,i,j,id;45     char op[5];46     double x1,y1,x2,y2;47     scanf("%d",&t);48     while(t--)49     {50         scanf("%d",&n);51         int cnt=0;52         for(i=0;i<n;i++)53         {54             scanf("%s",op);55             if(op[0]==‘P‘)56             {57                 scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);58                 L[++cnt].a=Point(x1,y1);L[cnt].b=Point(x2,y2);59                 f[cnt]=cnt;60                 for(j=1;j<=cnt-1;j++)61                     if(judge(L[cnt],L[j]))62                         Union(j,cnt);63             }64             else 65             {66                 int ans=0;scanf("%d",&id);67                 id=findset(id);68                 for(j=1;j<=cnt;j++)69                     if(findset(j)==id)70                         ans++;71                 printf("%d\n",ans);72             }73         }74         if(t) printf("\n");75     }76     return 0;77 }

 

hdu 1558 线段相交+并查集路径压缩


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