POJ2502 Subway(Floyd)
来源:互联网 发布:solidworks能编程吗 编辑:程序博客网 时间:2024/06/05 08:09
Description
You have just moved from a quiet Waterloo neighbourhood to a big, noisy city. Instead of getting to ride your bike to school every day, you now get to walk and take the subway. Because you don't want to be late for class, you want to know how long it will take you to get to school.
You walk at a speed of 10 km/h. The subway travels at 40 km/h. Assume that you are lucky, and whenever you arrive at a subway station, a train is there that you can board immediately. You may get on and off the subway any number of times, and you may switch between different subway lines if you wish. All subway lines go in both directions.
You walk at a speed of 10 km/h. The subway travels at 40 km/h. Assume that you are lucky, and whenever you arrive at a subway station, a train is there that you can board immediately. You may get on and off the subway any number of times, and you may switch between different subway lines if you wish. All subway lines go in both directions.
Input
Input consists of the x,y coordinates of your home and your school, followed by specifications of several subway lines. Each subway line consists of the non-negative integer x,y coordinates of each stop on the line, in order. You may assume the subway runs in a straight line between adjacent stops, and the coordinates represent an integral number of metres. Each line has at least two stops. The end of each subway line is followed by the dummy coordinate pair -1,-1. In total there are at most 200 subway stops in the city.
Output
Output is the number of minutes it will take you to get to school, rounded to the nearest minute, taking the fastest route.
Sample Input
0 0 10000 10000 200 5000 200 7000 200 -1 -1 2000 600 5000 600 10000 600 -1 -1
Sample Output
21
输入确实是坑。。我在本地都没有办法测试数据 硬是wa了数十次
floyd dijkstra spfa均可做
我觉得这题比较难的是建图。。 其实是因为自己不会建。。。
AC代码:
#include "iostream"#include "cstdio"#include "cstring"#include "algorithm"#include "cmath"using namespace std;const int MAXN = 305;struct node{/* data */double x, y;}point[MAXN];double map[MAXN][MAXN], subv, walkv;int n;double dist(node a, node b){return sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y));}int main(int argc, char const *argv[]){for(int i = 0; i < MAXN; ++i)for(int j = 0; j < MAXN; ++j)map[i][j] = -1;subv = 40 * 1000.0 / 60;walkv = 10 * 1000.0 / 60;for(int i = 0; i < 2; ++i)scanf("%lf%lf", &point[i].x, &point[i].y);n = 2;node last, now;last.x = last.y = -1;while(scanf("%lf%lf", &now.x, &now.y) != EOF) {if(!(now.x == -1 && now.y == -1)) {point[n++] = now;if(!(last.x == -1 && last.y == -1))map[n - 1][n - 2] = map[n - 2][n - 1] = dist(point[n - 1], point[n - 2]) / subv;}last = now;}for(int i = 0; i < n; ++i)for(int j = 0; j < n; ++j)if(map[i][j] == -1) map[i][j] = dist(point[i], point[j]) / walkv;for(int k = 0; k < n; ++k)for(int i = 0; i < n; ++i)for(int j = 0; j < n; ++j)map[i][j] = min(map[i][j], map[i][k] + map[k][j]);printf("%.0f\n", map[0][1]);return 0;}
0 0
- POJ2502 Subway(Floyd)
- poj2502 - Subway
- poj2502 SubWay
- Poj2502 Subway
- poj2502 Subway
- POJ2502 subway
- poj2502 Subway最短路
- POJ2502 Subway flody
- POJ2502——Subway
- POJ2502 Subway 最短路
- poj2502 subway dijkstra
- POJ2502 subway(spfa)
- POJ2502 Subway(最短路径)
- Floyd算法 POJ2502
- POJ2502 Subway -DIJKSTRA最短路练习
- POJ2502 Subway(最短路Floyed)
- 【最短路】POJ2502 SUBWAY (spfa)
- POJ2502 subway(dijkstra以最短时间代替最短路)
- 有哪些高效但是不常见的Linux命令?
- 8月9日,牢骚第一阶段 暑假一期计划
- poj 1088 滑雪【记忆化搜索】
- UVA - 10972 RevolC FaeLoN (边双连通分量)
- 黑马程序员---OC autorelease
- POJ2502 Subway(Floyd)
- Android Studio添加插件(Genymotion)
- BaseAdapter优化深入分析
- 使用C#详解常用排序算法(一):概述
- 关于进程和线程比较好的一个解释
- 三级联动(省市县)
- js timestamp与datetime之间的相互转换
- d3d9 使用模板测试绘制出镜子特效
- POJ 1426 Find The Multiple