HDOJ1558 线段相交问题+并查集处理

Segment set

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5104 Accepted Submission(s): 1967

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

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才刚做计算几何的题没多久就遇到了并查集的内容，看来并查集真要好好掌握啊！！

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

const int maxn = 1005;
struct point{
    double x,y;
};
struct segment{
    point s,e;
}a[maxn];
int i,j,k,n,tot;
int f[maxn],ans[maxn];
char cc;

double product(point &a, point &b, point &c){
   double x1,y1,x2,y2;
   x1 = a.x - c.x; y1 = a.y - c.y;
   x2 = b.x - c.x; y2 = b.y - c.y;
   return (x1*y2 - x2*y1);
}

bool intersect(segment &a, segment &b){
    if (/*max(a.s.x,a.e.x)>=min(b.s.x,b.e.x) &&
        max(a.s.y,a.e.y)>=min(b.s.y,b.e.y) &&
        max(b.s.x,b.e.x)>=min(a.s.x,a.e.x) &&
        max(b.s.y,b.e.y)>=min(a.s.y,a.e.y) &&8*/
        product(a.s,a.e,b.s)*product(a.s,a.e,b.e)<=0 &&
        product(b.s,b.e,a.s)*product(b.s,b.e,a.e)<=0 ) return 1;
    return 0;
}

int Find(int x) {
     if (f[x]==x) return x;
     return f[x] = Find(f[x]);
}

void Merge(int x, int y){
    int fx = Find(x), fy = Find(y);
    if (fx!=fy) {
        f[fx] = fy;
        ans[fy] += ans[fx];
    }
}

int main(){
   int T;
   scanf("%d",&T);

   while (T--) {
       tot = 0;
       for (i=1; i<=1000; i++) ans[i] = 1;
       for (i=1; i<=1000; i++) f[i] = i;

       scanf("%d",&n);
       while (n--) {
          cin >> cc;
          if (cc=='P') {
              tot++;
              scanf("%lf %lf %lf %lf",&a[tot].s.x,&a[tot].s.y,&a[tot].e.x,&a[tot].e.y);
              for (i=1; i<=tot-1; i++)
                if (intersect(a[i],a[tot])) {
                        Merge(tot,i);
                  }
          }
          else {
             scanf("%d",&k);
             printf("%d\n",ans[Find(k)]);
          }
       }
       if (T) printf("\n");
   }
   return 0;
}