POJ 2502 最短路

【POJ 2502 最短路】的更多相关文章

http://poj.org/problem?id=2502 同一条地铁线上的站点相邻点间按照v2建边,然后所有点之间按照v1更新建边,所有的边都是双向边,both directions. 然后直接跑dij就好了,防止因为重复的站点多建了同样的点,把上限开到了500,AC. 一个小bug是因为数组开了500,然后初始化时访问了500导致越界,,,把其他位置的地址更改了检查了半天,衰,以后还是多开几十好点. #include<iostream> #include<cstdio> #i…

Subway POJ - 2502 最短路

题意:给出地铁线起点和终点坐地铁速度为v2 走路为v1 求起点到终点的最短距离 (答案需要四舍五入这里坑了好久) 拿给出的地铁站点和起点终点建边即可然后跑个迪杰斯特拉 #include<iostream> #include<cstdio> #include<cstring> #include<cmath> using namespace std; ; ; int n; +; double dist[maxn]; double cost[m…

POJ 2502 - Subway Dijkstra堆优化试水

做这道题的动机就是想练习一下堆的应用,顺便补一下好久没看的图论算法. Dijkstra算法概述 //从0出发的单源最短路 dis[][] = {INF} ReadMap(dis); for i = 0 -> n - 1 d[i] = dis[0][i] while u = GetNearest(1 .. n - 1, !been[]) been[u] = 1 for_each edge from u d[edge.v] = min(d[edge.v], d[u] + dis[u][edge.v]…

POJ 2502 Subway / NBUT 1440 Subway / SCU 2186 Subway（图论，最短距离）

POJ 2502 Subway / NBUT 1440 Subway / SCU 2186 Subway(图论,最短距离) 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.…

Heavy Transportation POJ 1797 最短路变形

Heavy Transportation POJ 1797 最短路变形题意原题链接题意大体就是说在一个地图上,有n个城市,编号从1 2 3 ... n,m条路,每条路都有相应的承重能力,然后让你求从编号为1的城市到编号为n的城市的路线中,最大能经过多重的车. 解题思路这个题可以使用最短路的思路,不过转移方程变了\(dis[j]=max(dis[j], min(dis[u], e[u][j]))\).这里dis[j]表示从标号为1的点到达编号为j的点的路径中,最小的承重能力,就像短板效应样…

POJ 2502 Subway （最短路）

Subway 题目链接: http://acm.hust.edu.cn/vjudge/contest/122685#problem/L 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 s…

POJ 2502 Subway-经过预处理的最短路

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…