2013
12-12

# Segment set

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.

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.

For each Q-command, output the answer. There is a blank line between test cases.

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

1
2
2
2
5

1 有n个操作，最后求有几个集合或者说是连通分量
2 对于输入一条线段我们就去前面找能够和它相交的线段，利用并查集进行合并并且更新rank数组，rank[x]数组保存的是以x为跟节点的集合的线段的数量
3 这一题难点就是线段的相交判断

#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;

const double eps = 1e-8;
const double INF = 1<<30;
const int MAXN = 1010;

struct Node{
double x1;
double y1;
double x2;
double y2;
};
Node node[MAXN];

int n;
int father[MAXN];
int rank[MAXN];

void init(int m){
for(int i = 0 ; i <= m ; i++){
father[i] = i;
rank[i] = 1;
}
}

int find(int x){
if(father[x] != x){
int fa = father[x];
father[x] = find(fa);
rank[x] += rank[fa];
}
return father[x];
}

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;
}
void solve(){
for(int i = 1 ; i < n ; i++){
if(inter(node[i] , node[n])){
int fx = find(i);
int fy = find(n);
if(fx != fy){
father[fy] = fx;
rank[fx] += rank[fy];
}
}
}
}

int main(){
int cas , m , k;
bool isFirst = true;
char c;
scanf("%d" , &cas);
while(cas--){
if(isFirst)
isFirst = false;
else
puts("");
n = 1;
scanf("%d%*c" , &m);
init(m);
while(m--){
c = getchar();
if(c == 'P'){
scanf("%lf" ,&node[n].x1);
scanf("%lf" ,&node[n].y1);
scanf("%lf" ,&node[n].x2);
scanf("%lf%*c" ,&node[n].y2);
if(node[n].x1 > node[n].x2){
swap(node[n].x1 , node[n].x2);
swap(node[n].y1 , node[n].y2);
}
solve();
n++;
}
else{
scanf("%d%*c" , &k);
int fa = find(k);
printf("%d\n" , rank[fa]);
}
}
}
return 0;
}