这里介绍怎么求k短路 A*搜索 估价函数f[i]=g[i]+h[i]; 在这里g[i]表示到达点i当前路径长,h[i]表示点i到达终点的最短距离 在搜索中,每次都取队列估价函数值最小的点,然后把它所能到达的点更新进入队列 显然这需要一个优先队列来维护(heap) 当终点第k次出队时,当前路径长度就是k短路 ; type link=^node;      node=record        po,len:longint;        next:link;      end;      poin…

POJ3255 题意:给定一个图,求从1到n的次短路 分析:我们需要在dijkstra上作出一些修改,首先,到某个顶点v的次短路要么是到其他某个顶点u的最短路在加上u到v的边,要么是到v的次短路再加上u到v的边,因此我们需要记录的是最短和次短路. #include <iostream> #include <cstdio> #include <cstring> #include <string> #include <vector> #include…

题目链接: https://vjudge.net/problem/POJ-3255 题目大意: 给无向图,求1到n的次短路长度 思路: 由于边数较多,应该使用dijkstra的队列优化 用d数组存储最短路,用d2数组存储次短路,每次更新的时候,先松弛更新最短路,如果松弛更新成功,把之前的最短路取出,再和次短路比较,更新次短路.每次更新两个数组 #include<iostream> #include<vector> #include<queue> #include<…

题目传送门 Roadblocks Description Bessie has moved to a small farm and sometimes enjoys returning to visit one of her best friends. She does not want to get to her old home too quickly, because she likes the scenery along the way. She has decided to take…

题目: POJ3255 洛谷2865 分析: 这道题第一眼看上去有点懵-- 不过既然要求次短路,那估计跟最短路有点关系,所以就拿着优先队列优化的Dijkstra乱搞,搞着搞着就通了. 开两个数组:\(dis\)存最短路,\(dis2\)存次短路 在松弛的时候同时更新两个数组,要判断三个条件 (\(u\)是当前考虑的点,\(v\)是与\(u\)有边相连的点,\(d(u,v)\)表示从\(u\)到\(v\)的边长) 1.如果\(dis[v]>dis[u]+d(u,v)\),则更新\(dis[v]\)…

315. [POJ3255] 地砖RoadBlocks ★★★   输入文件:block.in   输出文件:block.out   简单对比时间限制:1 s   内存限制:128 MB Description Bessie has moved to a small farm and sometimes enjoys returning to visit one of her best friends. She does not want to get to her old home too q…

点这里看题目 3228K 485MS G++ 2453B 根据题意和测试用例知道这是一个求次短路径的题目.次短路径,就是比最短路径长那么一丢丢的路径,而题中又是要求从一点到指定点的次短路径,果断Dijkstra. R (1 ≤ R ≤ 100,000,N (1 ≤ N ≤ 5000) ,length D (1 ≤ D ≤ 5000),所以我用链式向前星方法存储,这个不知道的点这里(我转载别人的,讲的挺详细). 用一个二维数组dist[MAXN][2],去记录i->j的最短路径和次短路径,第二维是…

Roadblocks Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 13594   Accepted: 4783 Description Bessie has moved to a small farm and sometimes enjoys returning to visit one of her best friends. She does not want to get to her old home too…

一.题目 Description Bessie has moved to a small farm and sometimes enjoys returning to visit one of her best friends. She does not want to get to her old home too quickly, because she likes the scenery along the way. She has decided to take the second-s…

AOJ0189 http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=0189 题意 求某一办公室到其他办公室的最短距离. 多组输入,n表示n条关系,下面n次每次输入 x y d表示x到y的距离是d.需要注意的是n没有给定,需要根据输入来求. 输出办公室的编号和距离. 思路 任意两点之间的最短距离用floyd算法比较合适. 代码 #include <iostream> #include <cstdio> #include…