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

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

 

题意：相交的直线算是一个集合，问低n条直线的集合有多少条直线

题解：直线相交模板+并查集多开一个num初值为1的数组，将其合并可以了

#include<stdio.h>#include<string.h>#define eps 1e-6int 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;}