POJ1127 Jack Straws

题目链接：http://poj.org/problem?id=1127

Jack Straws

题目大意：给你n个线段，然后再给你m组数据，每组包含a，b两个线段，问你a，b是否CONNECTED ，（如果间接相交（即1交2,2交3,1虽然没有交3，但是也算相交），也算CONNECTED ）。

解题思路：将相交于同一条直线的放入同一个并查集里面，然后查找即可。

#include <iostream>
#include <cstdio>
#include <algorithm>
#define eps 1e-10
using namespace std;

struct point
{
    int x,y;
};

struct line
{
    point a,b;
}l[15];

int pa[15];

bool inter(point a,point b,point c,point d)
{
    if(min(a.x,b.x) > max(c.x,d.x) || min(a.y,b.y) > max(c.y,d.y) || min(c.x,d.x) > max(a.x,b.x) || min(c.y,d.y) > max(a.y,b.y))
       return 0;
    double h,i,j,k;
    h = (b.x-a.x) * (c.y-a.y) - (b.y-a.y) * (c.x-a.x);
    i = (b.x-a.x) * (d.y-a.y) - (b.y-a.y) * (d.x-a.x);
    j = (d.x-c.x) * (a.y-c.y) - (d.y-c.y) * (a.x-c.x);
    k = (d.x-c.x) * (b.y-c.y) - (d.y-c.y) * (b.x-c.x);
    return h*i <= eps && j*k <= eps;
}

int findset(int x)
{
    if(x == pa[x])
        return pa[x];
    else
        return pa[x]=findset(pa[x]);
}

int main()
{
    int n;
    while(scanf("%d",&n),n)
    {
        int i,j,t1,t2,x,y;
        for(i = 0; i <= n; i++)
            pa[i] = i;
        for(i = 1; i <= n; i++)
            scanf("%d%d%d%d",&l[i].a.x,&l[i].a.y,&l[i].b.x,&l[i].b.y);
        for(i = 1; i <= n; i++)
        {
            for(j = i+1; j <= n; j++)
            {
                if(inter(l[i].a,l[i].b,l[j].a,l[j].b))
                {
                    x = findset(i);
                    y = findset(j);
                    if(x != y)
                        pa[x] = y;
                }
            }
        }
        for(i = 1; i <= n; i++)
            pa[i] = findset(i);
        while(scanf("%d%d",&t1,&t2),t1&&t2)
        {
            if(pa[t1] == pa[t2])
                printf("CONNECTED\n");
            else
                printf("NOT CONNECTED\n");
        }
    }
    return 0;
}