POJ 1556 The Doors （计算几何判断线段相交+最短路）

题目链接：http://poj.org/problem?id=1556

题意：

1
5 4 6 7 8
2
4 2 7 8 9
7 3 4.5 6 7
-1

房间里有n堵墙，每面墙上有两扇门，求从房间最左端中点到最右端中点的最短路径。

第一行n表示有n面墙，下面n行每行第一个数为该墙的x坐标，后面4个数为四个点的y坐标。

这题wa了好久。。最后发现居然是floyd写错了。。。。。。。。。。。。

先把所有顶点存到图里，然后跑一遍floyd即可。

编号与输入数据的对应关系为 (i - 1) * 4 + j + 1，加上起点和终点一共4 * n + 1个点 

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>using namespace std;const double inf = 1e2;double x[25], y[25][4];  //读入数据 double m[100][100]; //图 double sx = 0, sy = 5;double ex = 10, ey = 5;const double eps = 1e-8;int n;int sgn(double x) {if(fabs(x) < eps)return 0;if(x < 0)return -1;  else return 1; }struct Point {  double x,y;Point(){}  Point(double _x,double _y) {   x = _x;y = _y;  }  Point operator -(const Point &b)const {   return Point(x - b.x,y - b.y);  }  double operator ^(const Point &b)const {   return x*b.y - y*b.x;  }double operator *(const Point &b)const {   return x*b.x + y*b.y;  }};struct Line {  Point s,e; Line(){}  Line(Point _s,Point _e)  {   s = _s;e = _e;  }};double dist(Point a,Point b) {     return sqrt((a-b)*(a-b)); }bool inter(Line l1,Line l2) {   //判断线段相交 return     max(l1.s.x,l1.e.x) >= min(l2.s.x,l2.e.x) &&     max(l2.s.x,l2.e.x) >= min(l1.s.x,l1.e.x) &&     max(l1.s.y,l1.e.y) >= min(l2.s.y,l2.e.y) &&     max(l2.s.y,l2.e.y) >= min(l1.s.y,l1.e.y) &&     sgn((l2.s-l1.e)^(l1.s-l1.e))*sgn((l2.e-l1.e)^(l1.s-l1.e)) <= 0 &&     sgn((l1.s-l2.e)^(l2.s-l2.e))*sgn((l1.e-l2.e)^(l2.s-l2.e)) <= 0; }//编号与输入数据的对应关系为 (i - 1) * 4 + j + 1，加上起点和终点一共4 * n + 1个点 void mk_map(double x1, double y1, int index, int indexm) {  //index表示此点的x坐标，indexm表示此点在图中的编号 int i, j, k;for(i = index + 1; i <= n; i++) { //遍历此顶点后面的墙上的所有顶点 for(j = 0; j < 4; j++) {Point s1(x1, y1), e1(x[i], y[i][j]);int flag = 1;for(k = index + 1; k < i; k++) {  //判断是否可达，即连线是否能与门相交 Point s2(x[k], y[k][0]), e2(x[k], y[k][1]);Point s3(x[k], y[k][2]), e3(x[k], y[k][3]);Line l1(s1, e1), l2(s2, e2), l3(s3, e3);if(!inter(l1, l2) && !inter(l1, l3)) {flag = 0;break;}}if(flag) m[indexm][(i - 1) * 4 + j + 1] = dist(s1, e1);  //能则更新矩阵 }}Point s1(x1, y1), e1(ex, ey);int flag = 1;for(k = index + 1; k <= n; k++) {  //最后判断是否能到达终点 Point s2(x[k], y[k][0]), e2(x[k], y[k][1]);Point s3(x[k], y[k][2]), e3(x[k], y[k][3]);Line l1(s1, e1), l2(s2, e2), l3(s3, e3);if(!inter(l1, l2) && !inter(l1, l3)) {flag = 0;break;}}if(flag) m[indexm][4 * n + 1] = dist(s1, e1);}int main() {while(~scanf("%d", &n)) {if(n == -1) break;int i, j, k;for(i = 0; i < 100; i++) {for(j = 0; j < 100; j++) {m[i][j] = (i == j ? 0 : inf); //图的初始化 }}for(i = 1; i <= n; i++) {scanf("%lf %lf %lf %lf %lf", x + i, &y[i][0], &y[i][1], &y[i][2], &y[i][3]);}mk_map(0, 5, 0, 0);  // 初始化起始点到所有可达点的边 for(i = 1; i <= n; i++) {for(j = 0; j < 4; j++) {mk_map(x[i], y[i][j], i, (i - 1) * 4 + j + 1); // 初始化所有门的顶点到其后可达点的边 }}//floyd for(k = 0; k <= 4 * n + 1; k++) {for(i = 0; i <= 4 * n + 1; i++) {for(j = 0; j <= 4 * n + 1; j++) {if(m[i][j] > m[i][k] + m[k][j]) m[i][j] = m[i][k] + m[k][j];}}}printf("%.2f\n", m[0][4 * n + 1]);}return 0;}