hdu1558 Segment set

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

这题可以用并查集做，主要是要判断两条线段是不是相交，我的方法是记两条线段的端点是p1(x1,y1),p2(x2,y2)以及p3(x3,y3),p4(x4,y4),分别看4个端点有没有在另一条线段上，如果有，那么相交，如果没有，那么再用叉积判断每条线段的两个端点是不是在另一条线段的两边,如果p1，p2在线段p3p4的两边而且p3，p4在线段p1p2的两边，那么p1p2和p3p4相交，否则不相交。那么怎么判断p1，p2是否在线段p3p4的两边呢，这里用到了叉积，第一次取p1,p3,p4,如果向量p1p3和向量p1p4的叉积是正的，那么p1p2在p3p4的顺时针方向，反之在逆时针方向。再取p2,p3,p4，方法相同，如果两个叉积符号不同，那么p1，p2在线段p3p4的两边,否则不在。

#include<stdio.h>
#include<string.h>
#include<math.h>
int pre[1006];
int find(int x){
    int r=x;
    while(r!=pre[r])r=pre[r];
    return r;
}

int shizhen(double x2,double y2,double x4,double y4,double x5,double y5){
    double x=x4-x2,y=y4-y2,xx=x5-x2,yy=y5-y2;
    if(x*yy-xx*y>0)return 1;
    else return -1;
}

int xianshang(double x2,double y2,double x3,double y3,double x4,double y4){
    if((x2-x3)*(y4-y2)-(y2-y3)*(x4-x2)!=0)return 0;
    if((x2<x3 && x2<x4) || (x2>x3 && x2>x4))return 0;
    return 1;
}

int panduan(double x2,double y2,double x3,double y3,double x4,double y4,double x5,double y5)
{
    if(xianshang(x2,y2,x4,y4,x5,y5) || xianshang(x3,y3,x4,y4,x5,y5) || xianshang(x4,y4,x2,y2,x3,y3) || xianshang(x5,y5,x2,y2,x3,y3))return 1;
    if((shizhen(x2,y2,x4,y4,x5,y5)*shizhen(x3,y3,x4,y4,x5,y5)<0)  && (shizhen(x4,y4,x2,y2,x3,y3)*shizhen(x5,y5,x2,y2,x3,y3)<0))return 1;
    return 0;
}

int main()
{
    int T,n,m,i,j,a,t1,xianduan[1006],num[1005],num1,h;
    double x2[1006],x3[1006],y2[1006],y3[1006];
    char s[10];
    scanf("%d",&T);
    for(h=1;h<=T;h++)
    {
        scanf("%d",&n);
        num1=0;
        for(i=1;i<=n;i++){
            pre[i]=i;num[i]=0;
        }

        for(i=1;i<=n;i++){
            scanf("%s",s);
            if(s[0]=='P'){
                num1++;
                scanf("%lf%lf%lf%lf",&x2[num1],&y2[num1],&x3[num1],&y3[num1]);
                num[num1]=1;
                for(j=1;j<=num1-1;j++){
                    if(panduan(x2[j],y2[j],x3[j],y3[j],x2[num1],y2[num1],x3[num1],y3[num1])){
                        t1=find(j);
                        if(num1!=t1){
                            pre[t1]=num1;num[num1]+=num[t1];
                        }
                        else continue;
                    }
                    //printf("%d ",panduan(x2[j],y2[j],x3[j],y3[j],x2[num1],y2[num1],x3[num1],y3[num1]));
                }
                //printf("\n");
            }
            else if(s[0]=='Q'){
                scanf("%d",&a);
                t1=find(a);
                printf("%d\n",num[t1]);
            }
        }
        if(h!=T)printf("\n");
        /*for(i=1;i<=5;i++){
            t1=find(i);
            printf("%d ",num[t1]);
        }
        printf("\n");*/
    }
    return 0;
}