hdu 1558 Segment set (並查集)

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Segment setTime Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3907    Accepted Submission(s): 1471


Problem DescriptionA 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.

 
InputIn 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.
 
OutputFor 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

 
AuthorLL 

題目大意：在一個平面直角座標系裡面，通過P操作不斷的加入線段，如果兩個線段有相交，就表明他們是一個集合裡面的。Q操作詢問當前情況下第k條線段所在的集合裡面有幾條線段。

並查集的題目，但是我覺得主要考幾何。我開始可以想到，通過判斷兩條線段是否有交點，如果有就放在一個集合裡面。這麼想的確很簡單，但是做起來真的十分麻煩。。

如果對於兩條線段，可以通過簡單計算得到兩者的交點x0=(b2-b1)/(k1-k2)，還有y0。那麼我只要判斷x0,y0是否線上段相交的地方即可。但是還要注意，這個交點是從k1,k2得到的。所以如果k1,k2不存在，又要分情況討論。

以下是My Code，感覺好像還有遺漏的地方，雖然的確是AC了。

#include<stdio.h>#include<string.h>int p[10000],sum[10000];double x1[1005],x2[1005],y1[1005],y2[1005];void init(int x){    int i;    for(i=0;i<=x;i++)    p[i]=i;    for(i=0;i<=x;i++)    sum[i]=1;}int findroot(int x){    int r=x;    while(r!=p[r])    r=p[r];    int i,j;    i=x;    while(i!=r)    {        j=p[i];        p[i]=r;        i=j;    }    return r;}void merge(int x,int y){    int fx=findroot(x);    int fy=findroot(y);    if(fx!=fy){        p[fx]=fy;        sum[fy]+=sum[fx];            }}double jiaodian(double x1,double y1,double x2,double y2,double x3,double y3,double x4,double y4){               if(x1==x2&&x3!=x4){             //k1不存在,k2存在         double k2=(y3-y4)/(x3-x4);        double y=k2*(x1-x3)+y3;        if((y>=y1&&y<=y2)||(y>=y2&&y<=y1))return 1;        else return 0;    }   else if(x3==x4&&x1!=x2){           //k2不存在，k1存在        double k1=(y1-y2)/(x1-x2);        double y=k1*(x3-x1)+y1;        if((y>=y3&&y<=y4)||(y>=y4&&y<=y3))return 1;        else return 0;    }   else if(x1==x2&&x3==x4){        if(x1==x3&&((y1>=y3&&y1<=y4)||(y1>=y4&&y1<=y3)||(y2>=y4&&y2<=y3)||(y2>=y3&&y2<=y4)))return 1;        else return 0;    }      double k1=(y1-y2)/(x1-x2);     double k2=(y3-y4)/(x3-x4);    double b1=(x1*y2-x2*y1)/(x1-x2);    double b2=(x3*y4-x4*y3)/(x3-x4);    double x=(b2-b1)/(k1-k2);    double y=k1*(x-x1)+y1;    if(((x>=x1&&x<=x2)||(x>=x2&&x<=x1))&&((y>=y1&&y<=y2)||(y>=y2&&y<=y1))&&    ((x>=x3&&x<=x4)||(x>=x4&&x<=x3))||((y>=y3&&y<=y4)&&(y>=y4&&y<=y3)))return 1;     return 0;}void isconnect(int x){    int i;    for(i=1;i<=x;i++)  {            if(jiaodian(x1[i],y1[i],x2[i],y2[i],x1[x],y1[x],x2[x],y2[x])){merge(i,x);}  }  return ;}int main(){    int t,n,i,j,k,m,cnt,q;         char c[10];    scanf("%d",&t);    while(t--)    {        q=1;        scanf("%d",&n);        init(n);                cnt=1;        for(i=1;i<=n;i++)        {            scanf("%s",c);            if(c[0]=='P')            {            scanf("%lf%lf%lf%lf",&x1[q],&y1[q],&x2[q],&y2[q]);                    if(i>1){             isconnect(q);                             }            q++;            }            if(c[0]=='Q'){                scanf("%d",&k);                int s=findroot(k);                cnt=sum[s];                printf("%d\n",cnt);                        }                    }            if(t>0)printf("\n");    }    return 0;}

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hdu 1558 Segment set (並查集)