hdu 1558 Segment set（判线段相交模版）

Segment set

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2644    Accepted Submission(s): 994

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
并查集水题，这题要判线段相交。
AC代码：
#include <iostream>#include <cstring>#include <string>#include <cstdio>#include <algorithm>#include <queue>#include <map>#include <cmath>#include <vector>#include <cstdlib>#define eps 1e-8using namespace std;const int MAX=1005;int father[MAX],rank[MAX];struct node{    double x1,y1;    double x2,y2;} line[1010];int max(int a,int b){    return a>b?a:b;}int min(int a,int b){    return a<b?a:b;}double multiply1(node a,node b){    return (a.x1-a.x2)*(b.y1-a.y1)-(a.y1-a.y2)*(b.x1-a.x1);}double multiply2(node a,node b){    return (a.x1-a.x2)*(b.y2-a.y1)-(a.y1-a.y2)*(b.x2-a.x1);}bool inter(node a,node b){    if(max(a.x1,a.x2)>=min(b.x1,b.x2)&&            max(b.x1,b.x2)>=min(a.x1,a.x2)&&            max(a.y1,a.y2)>=min(b.y1,b.y2)&&            max(b.y1,b.y2)>=min(a.y1,a.y2)&&            multiply1(a,b)*multiply2(a,b)<=eps&&            multiply1(b,a)*multiply2(b,a)<=eps)        return true;    else return false;}int Find_set(int n){    if(n!=father[n])        father[n]=Find_set(father[n]);    return father[n];}void Union(int a,int b){    a=Find_set(a);    b=Find_set(b);    if(a==b) return;    father[b]=a;    rank[a]+=rank[b];}void init(){    for(int i=0; i<MAX; i++)    {        father[i]=i;        rank[i]=1;    }}int main(){    int t,n,m;    char s[2];    scanf("%d",&t);    while(t--)    {        init();        int cnt=1;        scanf("%d",&n);        while(n--)        {            scanf("%s",s);            if(s[0]=='P')            {                scanf("%lf%lf%lf%lf",&line[cnt].x1,&line[cnt].y1,&line[cnt].x2,&line[cnt].y2);                for(int i=1; i<cnt; i++)                    if(inter(line[i],line[cnt]))                        Union(i,cnt);                cnt++;            }            else            {                scanf("%d",&m);                int root=Find_set(m);                printf("%d\n",rank[root]);            }        }        if(t>0)        printf("\n");    }    return 0;}