poj1127 Jack Straws(线段相交+并查集)

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Jack Straws

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 3512		Accepted: 1601

Description

In the game of Jack Straws, a number of plastic or wooden "straws" are dumped on the table and players try to remove them one-by-one without disturbing the other straws. Here, we are only concerned with if various pairs of straws are connected by a path of touching straws. You will be given a list of the endpoints for some straws (as if they were dumped on a large piece of graph paper) and then will be asked if various pairs of straws are connected. Note that touching is connecting, but also two straws can be connected indirectly via other connected straws.
Input

Input consist multiple case,each case consists of multiple lines. The first line will be an integer n (1 < n < 13) giving the number of straws on the table. Each of the next n lines contain 4 positive integers,x1,y1,x2 and y2, giving the coordinates, (x1,y1),(x2,y2) of the endpoints of a single straw. All coordinates will be less than 100. (Note that the straws will be of varying lengths.) The first straw entered will be known as straw #1, the second as straw #2, and so on. The remaining lines of the current case(except for the final line) will each contain two positive integers, a and b, both between 1 and n, inclusive. You are to determine if straw a can be connected to straw b. When a = 0 = b, the current case is terminated.

When n=0,the input is terminated.

There will be no illegal input and there are no zero-length straws.
Output

You should generate a line of output for each line containing a pair a and b, except the final line where a = 0 = b. The line should say simply "CONNECTED", if straw a is connected to straw b, or "NOT CONNECTED", if straw a is not connected to straw b. For our purposes, a straw is considered connected to itself.
Sample Input

7
1 6 3 3
4 6 4 9
4 5 6 7
1 4 3 5
3 5 5 5
5 2 6 3
5 4 7 2
1 4
1 6
3 3
6 7
2 3
1 3
0 0

2
0 2 0 0
0 0 0 1
1 1
2 2
1 2
0 0

0

Sample Output

CONNECTED
NOT CONNECTED
CONNECTED
CONNECTED
NOT CONNECTED
CONNECTED
CONNECTED
CONNECTED
CONNECTED

这题还是比较简单的，就是问两条线段是否直接或者间接的相连。注意考虑好有一段是重叠的情况即可

/**
* code generated by JHelper
* More info: https://github.com/AlexeyDmitriev/JHelper * @author xyiyy @https://github.com/xyiyy
*/

#include <iostream>
#include <fstream>

//#####################
//Author:fraud
//Blog: http://www.cnblogs.com/fraud/ //#####################
//#pragma comment(linker, "/STACK:102400000,102400000")
#include <iostream>
#include <sstream>
#include <ios>
#include <iomanip>
#include <functional>
#include <algorithm>
#include <vector>
#include <string>
#include <list>
#include <queue>
#include <deque>
#include <stack>
#include <set>
#include <map>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <climits>
#include <cctype>

using namespace std;
#define rep(X, N) for(int X=0;X<N;X++)
#define rep2(X, L, R) for(int X=L;X<=R;X++)

const int MAXN = 110;
//
// Created by xyiyy on 2015/8/8.
//

#ifndef JHELPER_EXAMPLE_PROJECT_UNIONFINDSET_HPP
#define JHELPER_EXAMPLE_PROJECT_UNIONFINDSET_HPP

int pa[MAXN], ra[MAXN];

void init(int n) {
rep(i, n + 1)pa[i] = i, ra[i] = 0;
}

int find(int x) {
if (pa[x] != x)pa[x] = find(pa[x]);
return pa[x];
}

int unite(int x, int y) {
x = find(x);
y = find(y);
if (x == y)return 0;
if (ra[x] < ra[y])pa[x] = y;
else {
pa[y] = x;
if (ra[x] == ra[y])ra[x]++;
}
return 1;
}

bool same(int x, int y) {
return find(x) == find(y);
}

#endif //JHELPER_EXAMPLE_PROJECT_UNIONFINDSET_HPP

//
// Created by xyiyy on 2015/8/10.
//

#ifndef JHELPER_EXAMPLE_PROJECT_P_HPP
#define JHELPER_EXAMPLE_PROJECT_P_HPP

const double EPS = 1e-9;

class P {
public:
double x, y;

P() { }

P(double _x, double _y) {
x = _x;
y = _y;
}

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

P  operator+(const P &p) {
return P(add(x, p.x), add(y, p.y));
}

P operator-(const P &p) {
return P(add(x, -p.x), add(y, -p.y));
}

P operator*(const double &d) {
return P(x * d, y * d);
}

P operator/(const double &d) {
return P(x / d, y / d);
}

double det(P p) {
return add(x * p.y, -y * p.x);
}

//线段相交判定
bool crsSS(P p1, P p2, P q1, P q2) {
if (max(p1.x, p2.x) + EPS < min(q1.x, q2.x))return false;
if (max(q1.x, q2.x) + EPS < min(p1.x, p2.x))return false;
if (max(p1.y, p2.y) + EPS < min(q1.y, q2.y))return false;
if (max(q1.y, q2.y) + EPS < min(p1.y, p2.y))return false;
/*(if((p1 - p2).det(q1 - q2) == 0){
return (on_seg(p1,p2,q1) || on_seg(p1,p2,q2) || on_seg(q1,q2,p1) || on_seg(q1,q2,p2));
}else{
P r = intersection(p1,p2,q1,q2);
return on_seg(p1,p2,r) && on_seg(q1,q2,r);

}*/
return (p2 - p1).det(q1 - p1) * (p2 - p1).det(q2 - p1) <= 0
&& (q2 - q1).det(p1 - q1) * (q2 - q1).det(p2 - q1) <= 0;
}

//直线和直线的交点
/*P isLL(P p1,P p2,P q1,P q2){
double d = (q2 - q1).det(p2 - p1);
if(sig(d)==0)return NULL;
return intersection(p1,p2,q1,q2);
}*/

//四点共圆判定
/*bool onC(P p1,P p2,P p3,P p4){
P c = CCenter(p1,p2,p3);
if(c == NULL) return false;
return add((c - p1).abs2(), -(c - p4).abs2()) == 0;
}*/

//三点共圆的圆心
/*P CCenter(P p1,P p2,P p3){
//if(disLP(p1, p2, p3) < EPS)return NULL;//三点共线
P q1 = (p1 + p2) * 0.5;
P q2 = q1 + ((p1 - p2).rot90());
P s1 = (p3 + p2) * 0.5;
P s2 = s1 + ((p3 - p2).rot90());
return isLL(q1,q2,s1,s2);
}*/

};

#endif //JHELPER_EXAMPLE_PROJECT_P_HPP

class poj1127 {
public:
void solve(std::istream &in, std::ostream &out) {
int n;
P *p = new P[110];
P *q = new P[110];
while (in >> n && n) {
init(n + 5);
rep2(i, 1, n) {
in >> p[i].x >> p[i].y >> q[i].x >> q[i].y;
}
rep2(i, 1, n) {
rep2(j, 1, n) {
if (p[i].crsSS(p[i], q[i], p[j], q[j]))unite(i, j);
}
}
int u, v;
while (in >> u >> v && (u && v)) {
if (same(u, v))out << "CONNECTED" << endl;
else out << "NOT CONNECTED" << endl;
}
}
}
};

int main() {
std::ios::sync_with_stdio(false);
std::cin.tie(0);
poj1127 solver;
std::istream &in(std::cin);
std::ostream &out(std::cout);
solver.solve(in, out);
return 0;
}

代码君