POJ - 1127 Jack Straws(几何)
题意:桌子上放着n根木棍,已知木棍两端的坐标。给定几对木棍,判断每对木棍是否相连。当两根木棍之间有公共点或可以通过相连的木棍间接的连在一起,则认为是相连的。
分析:
1、若线段i与j平行,且有部分重合,则相连。否则,求直线i与直线j交点,再判断该交点是否在两线段上,以确定是否相连。
2、flod整理一下所有的关系。
3、判断点是否在线段上:(线段i,点t)
(1)外积(l[i] - t) × (r[i] - t) = 0, 可判断点x是否在直线i上(两向量叉乘为0,两向量平行)
(2)内积(l[i] - x) · (r[i] - x) <= 0, 可判断点x是否落在l[i]与r[i]之间。(以x为顶点,l[i] - x和r[i] - x为边的角a,当内积<0----a = 180°,当内积=0,x与l[i]重合或x与r[i]重合)。
4、求两直线交点:(直线i,直线j)
(1)直线i上的某点t可表示为l[i] + w(r[i] - l[i]),w为系数
(2)点t在直线j上可表示为(l[j] - t) × (r[j] - t) = 0,由此可算出系数w,但由于t的表达式过于复杂,所以改用(r[j] - l[j]) × (t - l[j]) = 0来判断点t是否在直线j上
解得,w = ((r[j] - l[j]) × (l[j] - l[i])) / ((r[j] - l[j]) × (r[i] - l[i]))。
(3)将求出的w代入t,此时t为直线i与j的交点,但该交点不一定在线段i与j上,所以要判断后才可确定线段i与j是否相连。
#pragma comment(linker, "/STACK:102400000, 102400000") #include<cstdio> #include<cstring> #include<cstdlib> #include<cctype> #include<cmath> #include<iostream> #include<sstream> #include<iterator> #include<algorithm> #include<string> #include<vector> #include<set> #include<map> #include<stack> #include<deque> #include<queue> #include<list> #define Min(a, b) ((a < b) ? a : b) #define Max(a, b) ((a < b) ? b : a) const double eps = 1e-10; inline int dcmp(double a, double b){ if(fabs(a - b) < eps) return 0; return a > b ? 1 : -1; } typedef long long LL; typedef unsigned long long ULL; const int INT_INF = 0x3f3f3f3f; const int INT_M_INF = 0x7f7f7f7f; const LL LL_INF = 0x3f3f3f3f3f3f3f3f; const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f; const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1}; const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1}; const int MOD = 1e9 + 7; const double pi = acos(-1.0); const int MAXN = 13 + 10; const int MAXT = 10000 + 10; using namespace std; int vis[MAXN][MAXN]; int N; double add(double a, double b){//考虑误差的加法运算 if(abs(a + b) < eps * (abs(a) + abs(b))) return 0; return a + b; } struct Point{ double x, y; void read(){ scanf("%lf%lf", &x, &y); } Point(){} Point(double xx, double yy):x(xx), y(yy){} Point operator + (Point p){ return Point(add(x, p.x), add(y, p.y)); } Point operator - (Point p){ return Point(add(x, -p.x), add(y, -p.y)); } Point operator * (double t){ return Point(x * t, y * t); } double dot(Point p){//内积 return add(x * p.x, y * p.y); } double det(Point p){//外积 return add(x * p.y, -y * p.x); } }l[MAXN], r[MAXN];//线段的左右端点 bool onsegment(Point A, Point B, Point x){//判断点x是否在以A,B为端点的线段上 return (A - x).det(B - x) == 0 && (A - x).dot(B - x) <= 0; } Point intersection(Point l1, Point r1, Point l2, Point r2){//计算直线1与直线2的交点(l1--直线1的左端点,r1--直线1的右端点) return l1 + (r1 - l1) * ((r2 - l2).det(l2 - l1) / (r2 - l2).det(r1 - l1)); } bool judge(Point l1, Point r1, Point l2, Point r2){ if(onsegment(l1, r1, l2)) return true; if(onsegment(l1, r1, r2)) return true; if(onsegment(l2, r2, l1)) return true; if(onsegment(l2, r2, r1)) return true; return false; } void solve(){ for(int i = 0; i < N; ++i){ vis[i][i] = 1; for(int j = 0; j < i; ++j){ if((l[i] - r[i]).det(l[j] - r[j]) == 0){//线段i与线段j平行 if(judge(l[i], r[i], l[j], r[j])){ vis[i][j] = vis[j][i] = 1;//两线段有部分重合,相连 } } else{ Point x = intersection(l[i], r[i], l[j], r[j]);//x为直线i与直线j的交点 vis[i][j] = vis[j][i] = onsegment(l[i], r[i], x) && onsegment(l[j], r[j], x);//需要判断该交点是否在两个线段上 } } } for(int k = 0; k < N; ++k){ for(int i = 0; i < N; ++i){ for(int j = 0; j < N; ++j){ vis[i][j] |= vis[i][k] && vis[k][j]; } } } } int main(){ while(scanf("%d", &N) == 1){ if(!N) return 0; memset(vis, 0, sizeof vis); for(int i = 0; i < N; ++i){ l[i].read(); r[i].read(); } solve(); int a, b; while(scanf("%d%d", &a, &b) == 2){ if(!a && !b) break; if(vis[a - 1][b - 1]){ printf("CONNECTED\n"); } else{ printf("NOT CONNECTED\n"); } } } return 0; }