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.
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
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 /** 2 * code generated by JHelper 3 * More info: https://github.com/AlexeyDmitriev/JHelper 4 * @author xyiyy @https://github.com/xyiyy 5 */ 6 7 #include <iostream> 8 #include <fstream> 9 10 //##################### 11 //Author:fraud 12 //Blog: http://www.cnblogs.com/fraud/ 13 //##################### 14 //#pragma comment(linker, "/STACK:102400000,102400000") 15 #include <iostream> 16 #include <sstream> 17 #include <ios> 18 #include <iomanip> 19 #include <functional> 20 #include <algorithm> 21 #include <vector> 22 #include <string> 23 #include <list> 24 #include <queue> 25 #include <deque> 26 #include <stack> 27 #include <set> 28 #include <map> 29 #include <cstdio> 30 #include <cstdlib> 31 #include <cmath> 32 #include <cstring> 33 #include <climits> 34 #include <cctype> 35 36 using namespace std; 37 #define rep(X, N) for(int X=0;X<N;X++) 38 #define rep2(X, L, R) for(int X=L;X<=R;X++) 39 40 const int MAXN = 110; 41 // 42 // Created by xyiyy on 2015/8/8. 43 // 44 45 #ifndef JHELPER_EXAMPLE_PROJECT_UNIONFINDSET_HPP 46 #define JHELPER_EXAMPLE_PROJECT_UNIONFINDSET_HPP 47 48 int pa[MAXN], ra[MAXN]; 49 50 void init(int n) { 51 rep(i, n + 1)pa[i] = i, ra[i] = 0; 52 } 53 54 int find(int x) { 55 if (pa[x] != x)pa[x] = find(pa[x]); 56 return pa[x]; 57 } 58 59 int unite(int x, int y) { 60 x = find(x); 61 y = find(y); 62 if (x == y)return 0; 63 if (ra[x] < ra[y])pa[x] = y; 64 else { 65 pa[y] = x; 66 if (ra[x] == ra[y])ra[x]++; 67 } 68 return 1; 69 } 70 71 bool same(int x, int y) { 72 return find(x) == find(y); 73 } 74 75 #endif //JHELPER_EXAMPLE_PROJECT_UNIONFINDSET_HPP 76 77 // 78 // Created by xyiyy on 2015/8/10. 79 // 80 81 #ifndef JHELPER_EXAMPLE_PROJECT_P_HPP 82 #define JHELPER_EXAMPLE_PROJECT_P_HPP 83 84 85 const double EPS = 1e-9; 86 87 class P { 88 public: 89 double x, y; 90 91 P() { } 92 93 P(double _x, double _y) { 94 x = _x; 95 y = _y; 96 } 97 98 double add(double a, double b) { 99 if (fabs(a + b) < EPS * (fabs(a) + fabs(b)))return 0; 100 return a + b; 101 } 102 103 P operator+(const P &p) { 104 return P(add(x, p.x), add(y, p.y)); 105 } 106 107 P operator-(const P &p) { 108 return P(add(x, -p.x), add(y, -p.y)); 109 } 110 111 P operator*(const double &d) { 112 return P(x * d, y * d); 113 } 114 115 P operator/(const double &d) { 116 return P(x / d, y / d); 117 } 118 119 double det(P p) { 120 return add(x * p.y, -y * p.x); 121 } 122 123 //线段相交判定 124 bool crsSS(P p1, P p2, P q1, P q2) { 125 if (max(p1.x, p2.x) + EPS < min(q1.x, q2.x))return false; 126 if (max(q1.x, q2.x) + EPS < min(p1.x, p2.x))return false; 127 if (max(p1.y, p2.y) + EPS < min(q1.y, q2.y))return false; 128 if (max(q1.y, q2.y) + EPS < min(p1.y, p2.y))return false; 129 /*(if((p1 - p2).det(q1 - q2) == 0){ 130 return (on_seg(p1,p2,q1) || on_seg(p1,p2,q2) || on_seg(q1,q2,p1) || on_seg(q1,q2,p2)); 131 }else{ 132 P r = intersection(p1,p2,q1,q2); 133 return on_seg(p1,p2,r) && on_seg(q1,q2,r); 134 135 }*/ 136 return (p2 - p1).det(q1 - p1) * (p2 - p1).det(q2 - p1) <= 0 137 && (q2 - q1).det(p1 - q1) * (q2 - q1).det(p2 - q1) <= 0; 138 } 139 140 //直线和直线的交点 141 /*P isLL(P p1,P p2,P q1,P q2){ 142 double d = (q2 - q1).det(p2 - p1); 143 if(sig(d)==0)return NULL; 144 return intersection(p1,p2,q1,q2); 145 }*/ 146 147 //四点共圆判定 148 /*bool onC(P p1,P p2,P p3,P p4){ 149 P c = CCenter(p1,p2,p3); 150 if(c == NULL) return false; 151 return add((c - p1).abs2(), -(c - p4).abs2()) == 0; 152 }*/ 153 154 //三点共圆的圆心 155 /*P CCenter(P p1,P p2,P p3){ 156 //if(disLP(p1, p2, p3) < EPS)return NULL;//三点共线 157 P q1 = (p1 + p2) * 0.5; 158 P q2 = q1 + ((p1 - p2).rot90()); 159 P s1 = (p3 + p2) * 0.5; 160 P s2 = s1 + ((p3 - p2).rot90()); 161 return isLL(q1,q2,s1,s2); 162 }*/ 163 164 165 }; 166 167 168 #endif //JHELPER_EXAMPLE_PROJECT_P_HPP 169 170 class poj1127 { 171 public: 172 void solve(std::istream &in, std::ostream &out) { 173 int n; 174 P *p = new P[110]; 175 P *q = new P[110]; 176 while (in >> n && n) { 177 init(n + 5); 178 rep2(i, 1, n) { 179 in >> p[i].x >> p[i].y >> q[i].x >> q[i].y; 180 } 181 rep2(i, 1, n) { 182 rep2(j, 1, n) { 183 if (p[i].crsSS(p[i], q[i], p[j], q[j]))unite(i, j); 184 } 185 } 186 int u, v; 187 while (in >> u >> v && (u && v)) { 188 if (same(u, v))out << "CONNECTED" << endl; 189 else out << "NOT CONNECTED" << endl; 190 } 191 } 192 } 193 }; 194 195 int main() { 196 std::ios::sync_with_stdio(false); 197 std::cin.tie(0); 198 poj1127 solver; 199 std::istream &in(std::cin); 200 std::ostream &out(std::cout); 201 solver.solve(in, out); 202 return 0; 203 }