计算几何-----》两直线是否相交

POJ 1127

  1 #include<set>
  2 #include<map>
  3 #include<stack>
  4 #include<bitset>
  5 #include<cmath>
  6 #include<string>
  7 #include<vector>
  8 #include<cstdio>
  9 #include<cstring>
 10 #include<iostream>
 11 #include<algorithm>
 12 #define pi acos(-1)
 13 #define close ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
 14 using namespace std;
 15 typedef long long ll;
 16 const int MAX_N=1000+50;
 17 const int INF=0x3f3f3f3f;
 18 const double EPS = 1e-10;
 19 ll mod = 1e9+7;
 20
 21
 22
 23 //考虑误差的加法运算
 24 double add(double a,double b){
 25     if(abs(a + b) < EPS * (abs(a) + abs(b))) return 0;
 26     return a+ b;
 27 }
 28
 29 //二维向量结构体
 30 struct P{
 31     double x,y;
 32     P(){}
 33     P(double x, double y) : x(x), y(y){
 34     }
 35     P operator + (P p){
 36         return P(add(x, p.x), add(y, p.y));
 37     }
 38     P operator - (P p){
 39         return P(add(x, -p.x), add(y, -p.y));
 40     }
 41     P operator * (double d){
 42         return P(x * d, y * d);
 43     }
 44     double dot(P p){  //内积
 45         return add(x * p.x, y * p.y);
 46     }
 47     double det(P p){ //外积
 48         return add(x * p.y, -y * p.x);
 49     }
 50 };
 51
 52 //判断点q是否在线段p1-p2上
 53 bool on_seg(P p1, P p2, P q){
 54     return (p1 - q).det(p2-q) == 0 && (p1 - q).dot(p2 - q) <= 0;
 55 }
 56
 57 //计算直线p1-p2与直线q1-q2的交点
 58 P intersection(P p1, P p2, P q1, P q2){
 59     return p1 + (p2 - p1) * ((q2 - q1).det(q1 - p1) / (q2 - q1).det(p2 - p1));
 60 }
 61
 62 int n;
 63 P p[MAX_N],q[MAX_N];
 64 int m;
 65 int a[MAX_N],b[MAX_N];
 66
 67 bool g[MAX_N][MAX_N];
 68
 69 void solve(){
 70     for(int i = 0; i < n; i++){
 71         g[i][i] = true;
 72         for(int j = 0; j < i; j++){
 73             //判断木棍i和木棍j是否有公共点
 74             if((p[i] - q[i]).det(p[j] - q[j]) == 0){
 75                 //平行时
 76                 g[i][j] = g[j][i] = on_seg(p[i], q[i], p[j])
 77                                 ||  on_seg(p[i], q[i], q[j])
 78                                 ||  on_seg(p[j], q[j], p[i])
 79                                 ||  on_seg(p[j], q[j], q[i]);
 80             }else{
 81                 //非平行时
 82                 P r = intersection(p[i], q[i], p[j], q[j]);
 83                 g[i][j] = g[j][i] = on_seg(p[i], q[i], r) && on_seg(p[j],q[j],r);
 84             }
 85         }
 86     }
 87     //通过Floyd_Warshall算法判断任意两点间是否相连
 88     for(int k = 0; k < n; k++){
 89         for(int i = 0; i< n; i++){
 90             for(int j = 0; j < n; j++){
 91                 g[i][j] |= g[i][k] && g[k][j];
 92             }
 93         }
 94     }
 95
 96     for(int i = 0; i < m; i++){
 97         puts(g[a[i] - 1][b[i] - 1] ? "CONNECTED" : "NOTCONNECTED");
 98     }
 99 }
100
101 int main(){
102     cin>>n;
103     for(int i = 0; i < n; i++){
104         cin>>p[i].x>>p[i].y>>q[i].x>>q[i].y;
105     }
106     solve();
107     int c, d;
108     while(~scanf("%d%d", &c, &d)) {
109         if(c==0 && d==0)    break;
110         if(g[c-1][d-1])
111             printf("CONNECTED\n");
112         else
113             printf("NOT CONNECTED\n");
114     }
115      return 0;
116 }
117
118 /*
119                 ********
120                ************
121                ####....#.
122              #..###.....##....
123              ###.......######              ###            ###
124                 ...........               #...#          #...#
125                ##*#######                 #.#.#          #.#.#
126             ####*******######             #.#.#          #.#.#
127            ...#***.****.*###....          #...#          #...#
128            ....**********##.....           ###            ###
129            ....****    *****....
130              ####        ####
131            ######        ######
132 ##############################################################
133 #...#......#.##...#......#.##...#......#.##------------------#
134 ###########################################------------------#
135 #..#....#....##..#....#....##..#....#....#####################
136 ##########################################    #----------#
137 #.....#......##.....#......##.....#......#    #----------#
138 ##########################################    #----------#
139 #.#..#....#..##.#..#....#..##.#..#....#..#    #----------#
140 ##########################################    ############
141 */

原文地址：https://www.cnblogs.com/Yinchen-One/p/9399449.html