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\(dis[v]\)…

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-sh…

Roadblocks Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 10098   Accepted: 3620 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…

题目大意:求图的严格次短路. 方法1: SPFA,同时求单源最短路径和单源次短路径.站在节点u上放松与其向量的v的次短路径时时,先尝试由u的最短路径放松,再尝试由u的次短路径放松(该两步并非非此即彼). 由u的最短路径放松: if(u->Dist + e->Weight < v->Dist) v->Dist2=v->Dist; //此处隐藏最短路放松.次短路不在此固定,Dist2可能在由u次短路放松时被放松得更短 if(u->Dist + e->Weight…

描述 求1到n的次最短路 开个\(dis[maxn][2]\)的储存距离的二维数组,0储存最短路,1储存次短路 初始化全为正无穷,\(dis[1][0]=0;\) 然后遍历更新时,先尝试更新最短路和次短路,不行就尝试更新次短路; 如果入队的距离已经大于次短路的距离,那么没必要继续,continue; #include <iostream> #include <cstring> #include <cstdio> #include <queue> #inclu…

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…

Roadblocks Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 7760   Accepted: 2848 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…