HDU_1558_SegmentSet

Segment set

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4100    Accepted Submission(s): 1550

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

 

Source

HDU 2006-12 Programming Contest

 

Recommend

LL

线段相交+并查集问题

问题问的是集合中的元素个数

因此合并操作的时候也要合并元素个数

#include <iostream>#include <stdio.h>#include <string.h>using namespace std;const int MN=1005;const double AC=1e-3;char con[2];double se[MN][4];           //输入的直线int senc[MN];               //直线的几何关系int sen[MN];                //集合中的直线个数int find(int n)             //查找{    if(senc[n]==n)        return n;    int t=senc[n];    return senc[n]=find(t);}double dt(double v1,double v2,double v3,double v4)       //行列式{    return v1*v4-v2*v3;}bool isx(int i,int j)                                      //判断相交{    double t1=dt(se[i][2]-se[i][0],se[j][0]-se[j][2],se[i][3]-se[i][1],se[j][1]-se[j][3]);    if(t1<=AC&&t1>=-AC)        return 0;    double t2=dt(se[j][0]-se[i][0],se[j][0]-se[j][2],se[j][1]-se[i][1],se[j][1]-se[j][3])/t1;    if(t2>1||t2<0)        return 0;    double t3=dt(se[i][2]-se[i][0],se[j][0]-se[i][0],se[i][3]-se[i][1],se[j][1]-se[i][1])/t1;    if(t3>1||t3<0)        return 0;    return 1;}void mer(int i,int j){    //cout<<i<<" "<<j<<endl;    int t1=find(i);    int t2=find(j);    if(t1==t2)        return;    sen[t2]=sen[t1]+sen[t2];   //合并元素个数    senc[t1]=t2;    //cout<<t2<<" "<<sen[t2]<<endl;}int main(){    int t;    int n;    int nn;    scanf("%d",&t);    while(t--)    {        int p=1;        scanf("%d",&n);        for(int i=1;i<=n;i++)         //重置        {            senc[i]=i;            sen[i]=1;        }        for(int i=1;i<=n;i++)        {            scanf("%s",con);            if(con[0]=='Q')            {                scanf("%d",&nn);                printf("%d\n",sen[find(nn)]);            }            else            {                scanf("%lf%lf%lf%lf",&se[p][0],&se[p][1],&se[p][2],&se[p][3]);                for(int i=1;i<p;i++)                {                    if(isx(i,p))                        mer(i,p);                }                p++;            }        }        if(t)            printf("\n");    }    return 0;}