hdu 1558 Segment set（并查集）

                                                       Segment set

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 2690    Accepted Submission(s): 1015

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
        

Author LL  

Source HDU 2006-12 Programming Contest  

Recommend LL  

題意：相交的直線算是一個集合，問低n條直線的集合有多少條直線

題解：直線相交模闆+并查集多開一個num初值為1的數組，将其合并可以了

#include<stdio.h>
#include<string.h>
#define eps 1e-6
int v[1005],num[1005];
struct segment{
    double x1,y1,x2,y2;
}s[1005];
double MAX(double a,double b)
{
    return a>b?a:b;
}
double MIN(double a,double b)
{
    return a<b?a:b;
}
double multiply1(int a,int b)
{
    return (s[a].x1-s[a].x2)*(s[b].y1-s[a].y1)-(s[a].y1-s[a].y2)*(s[b].x1-s[a].x1);
}
double multiply2(int a,int b)
{
    return (s[a].x1-s[a].x2)*(s[b].y2-s[a].y1)-(s[a].y1-s[a].y2)*(s[b].x2-s[a].x1);
}
int judge(int a,int b)
{
    if(MAX(s[a].x1,s[a].x2)>=MIN(s[b].x1,s[b].x2)&&
       MAX(s[b].x1,s[b].x2)>=MIN(s[a].x1,s[a].x2)&&
       MAX(s[a].y1,s[a].y2)>=MIN(s[b].y1,s[b].y2)&&
       MAX(s[b].y1,s[b].y2)>=MIN(s[a].y1,s[a].y2)&&
       multiply1(a,b)*multiply2(a,b)<=eps&&
       multiply1(b,a)*multiply2(b,a)<=eps)
       return 1;
    else return 0;
}
int f(int x)
{
    if(x==v[x]) return x;
    return v[x]=f(v[x]);
}
int main()
{
    int i,j,n,t,x,y,cou;
    char ch;

    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(i=0;i<=n;i++) v[i]=i,num[i]=1;
        for(cou=i=0;i<n;i++)
        {
            getchar();
            scanf("%c",&ch);
            if(ch=='P')
            {
                scanf("%lf%lf%lf%lf",&s[cou].x1,&s[cou].y1,&s[cou].x2,&s[cou].y2);
                for(j=0;j<cou;j++)
                {
                    if(judge(j,cou))
                    {
                        x=f(j),y=f(cou);
                        if(x==y) continue;
                        v[x]=y,num[y]+=num[x];
                    }
                }
                cou++;
            }
            else
            {
                scanf("%d",&x);
                printf("%d\n",num[f(x-1)]);
            }
        }
        if(t) printf("\n");
    }

    return 0;
}