poj 1127 Jack Straws 线段相交+并查集

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <queue>
#include <map>
#include <algorithm>
#include <set>
using namespace std;
#define MM(a) memset(a,0,sizeof(a))
typedef long long ll;
typedef unsigned long long ULL;
const int mod = 1000000007;
const double eps = 1e-10;
const int inf = 0x3f3f3f3f;
const int big=50000;
int max(int a,int b) {return a>b?a:b;};
int min(int a,int b) {return a<b?a:b;};
int r[15];
struct node{
   int  x,y;
}p;
struct Line{
   node a,b;
}line[15];
int neiji(node a,node b)
{
    return  a.x*b.x+a.y*b.y;
}

int jian(node a,node b,node c)
{
    return (b.x-a.x)*(c.y-a.y)-(c.x-a.x)*(b.y-a.y);
}

int findr(int i)
{
    if(i!=r[i])
        r[i]=findr(r[i]);
    return r[i];
}
void unite(int i,int j)
{
    int ri=findr(i),rj=findr(j);
    r[rj]=ri;
}
int solve(Line m,Line n)
{
    if(min(m.a.x,m.b.x)<=max(n.a.x,n.b.x)&&
       max(m.a.x,m.b.x)>=min(n.a.x,n.b.x)&&
       min(m.a.y,m.b.y)<=max(n.a.y,n.b.y)&&
       max(m.a.y,m.b.y)>=min(n.a.y,n.b.y)&&
       jian(m.a,m.b,n.a)*jian(m.a,m.b,n.b)<=0&&
       jian(n.a,n.b,m.a)*jian(n.a,n.b,m.b)<=0)
       return 1;
    return 0;
}//跨立实验判断线段相交
int main()
{
    int n;
    while(~scanf("%d",&n)&&n)
    {

        for(int i=1;i<=n;i++)
          scanf("%d %d %d %d",&line[i].a.x,&line[i].a.y,&line[i].b.x,&line[i].b.y);

        for(int i=1;i<=n;i++)
            r[i]=i;

        for(int i=1;i<=n;i++)
            for(int j=1;j<=n;j++)/*并查集检验线段的连通性，注意是两层循环，每个i都将直接与其相连的加入它当前所在的集合*/
           {
            if(findr(i)==findr(j))continue;
            if(solve(line[i],line[j]))
                 unite(i,j);
           }

        while(1)
        {
            int a,b;
            scanf("%d %d",&a,&b);
            if(!a||!b) break;
            if(findr(a)==findr(b))
                printf("CONNECTED\n");
            else
                printf("NOT CONNECTED\n");
        }
    }
    return 0;
}