poj 1127

//200k, 32ms
#include <iostream>
#include <cstdio>
#include <cmath>

using namespace std;

const int MAX_M = 10005;
const int MAX_N = 20;
const double EPS = 1e-10;

double add(double a, double b) {
    if(abs(a + b) < EPS * (abs(a) + abs(b)))    return 0;
    return a + b;
}

struct P {
    double x, y;
    P() {}
    P(double x, double y) : x(x), y(y) {}
    P operator + (P p) { return P(add(x, p.x), add(y, p.y)); }
    P operator - (P p) { return P(add(x, -p.x), add(y, -p.y)); }
    P operator * (double d) { return P(x * d, y * d); }
    double dot(P p) { return add(x * p.x, y * p.y); }
    double det(P p) { return add(x * p.y, -y * p.x); }
};

int n, m;
int a[MAX_M], b[MAX_M];
P p[MAX_N], q[MAX_N];

bool g[MAX_N][MAX_N];

//判断点q是否在线段p1-p2上 
bool on_seg(P p1, P p2, P q) {
    return (p1 - q).det(p2 - q) == 0 && (p1 - q).dot(p2 - q) <= 0;
}

//计算直线p1-p2与直线q1-12的交点 
P intersection(P p1, P p2, P q1, P q2) {
    return p1 + (p2 - p1 ) * ((q2 - q1).det(q1 - p1) / (q2 - q1).det(p2 - p1));
}

void solve() {
    for(int i=0; i<n; i++) {
        g[i][i] = true;
        for(int j=0; j<i; j++) {
            if((p[i] - q[i]).det(p[j] - q[j]) == 0) {
                g[i][j] = g[j][i] = on_seg(p[i], q[i], p[j])
                       || on_seg(p[i], q[i], q[j])
                       || on_seg(p[j], q[j], p[i])
                       || on_seg(p[j], q[j], q[i]);
            } else {
                P r = intersection(p[i], q[i], p[j], q[j]);
                g[i][j] = g[j][i] = on_seg(p[i], q[i], r) && on_seg(p[j], q[j], r);
            }
        }
    }
    //floyd-warshall
    for(int k=0; k<n; k++) {
        for(int i=0; i<n; i++) {
            for(int j=0; j<n; j++) {
                g[i][j] |= g[i][k] && g[k][j];
            }
        }
    }
}

int main() {

    freopen("in.txt", "r", stdin);

    while(~scanf("%d", &n)) {
        if(n == 0)  break;
        for(int i=0; i<n; i++) {
            scanf("%lf%lf%lf%lf", &p[i].x, &p[i].y, &q[i].x, &q[i].y);
        }
        solve();
        int c, d;
        while(~scanf("%d%d", &c, &d)) {
            if(c==0 && d==0)    break;
            if(g[c-1][d-1])
                printf("CONNECTED\n");
            else
                printf("NOT CONNECTED\n");
        }
    }

    fclose(stdin);

    return 0;
}

并查集：

//248k, 47ms
#include <iostream>
using namespace std;
/*--------- 并查集 模板 ------------*/
int UFS_NUM;    //并查集中元素总数 
typedef struct node{
    int data;    //节点对应的编号 
    int rank;    //节点对应秩 
    int parent;    //节点对应的双亲下标 
}UFSTree;        //并查集树的节点类型 
void MAKE_SET(UFSTree t[])    //初始化并查集树 
{
    int i;
    for(i=1;i<=UFS_NUM;i++){
        t[i].data = i;
        t[i].rank = 0;
        t[i].parent = i;
    }
}
int FIND_SET(UFSTree t[],int x)    //在x所在的子树中查找集合编号
{
    if(t[x].parent == x)
        return x;
    else 
        return FIND_SET(t,t[x].parent);
} 
void UNION(UFSTree t[],int x,int y)    //将x和y所在的子树合并
{
    x = FIND_SET(t,x);
    y = FIND_SET(t,y);
    if(t[x].rank > t[y].rank)
        t[y].parent = x;
    else{
        t[x].parent = y;
        if(t[x].rank==t[y].rank)
            t[y].rank++;
    }
} 

/*--------- 判断两线段相交 模板 ------------*/ 
const double eps=1e-10;
struct point { double x, y; };
double min(double a, double b) { return a < b ? a : b; }
double max(double a, double b) { return a > b ? a : b; }
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;
}

/*---------- 代码实现 -----------*/ 
struct line
{
    point p1;
    point p2;
};
int main()
{
    int n;
    UFSTree t[105]; 
    while(cin>>n){
        if(n==0) break;
        UFS_NUM = n;//确定并查集树中元素总数 
        MAKE_SET(t);    //初始化并查集 
        line l[15];
        for(int i=1;i<=n;i++)
            cin>>l[i].p1.x>>l[i].p1.y>>l[i].p2.x>>l[i].p2.y;
        for(int i=1;i<=n;i++)    //根据关系生成关系树 
            for(int j=1;j<=n;j++){
                if(i==j) continue;
                if(inter(l[i].p1,l[i].p2,l[j].p1,l[j].p2)){    //如果相交，有亲戚关系 
                    UNION(t,i,j);    //合并相关集合 
                }
            }
        int l1,l2;
        while(cin>>l1>>l2){
            if(l1==0 && l2==0)
                break;
            l1 = FIND_SET(t,l1);
            l2 = FIND_SET(t,l2);
            if(l1 == l2)
                cout<<"CONNECTED"<<endl;
            else
                cout<<"NOT CONNECTED"<<endl;
        }
    }
    return 0;
}