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…

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…

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…

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: 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\(dis[v]\)…

Roadblocks 直接翻译了 Descriptions Bessie搬到了一个新的农场,有时候他会回去看他的老朋友.但是他不想很快的回去,他喜欢欣赏沿途的风景,所以他会选择次短路,因为她知道一定有一条次短路.这个乡村有R(1<=R<=100000)条双向道路,每一条连接N(1<=N<=5000)个点中的两个.Bessie在1号节点,他的朋友家是n号节点Input第一行:两个整数N和R接下来R行:每行包含三个整数,A,B,D,表示一条连接A与B的长度为D的路径Output输出1到…

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…

Roadblocks http://poj.org/problem?id=3255 Time Limit: 2000MS   Memory Limit: 65536K       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 quick…

http://poj.org/problem?id=3255 同匈牙利游戏. 但是我发现了一个致命bug. 就是在匈牙利那篇,应该dis2单独if,而不是else if,因为dis2和dis1相对独立.有可能在前边两个if改了后还有更优的次短路. 所以,,wikioi那题太水,让我水过了.. #include <cstdio> #include <cstring> #include <cmath> #include <string> #include <…

题目链接 次短路模板题. 对每个点记录最短路和严格次短路,然后就是维护次值的方法了. 和这题一样. #include <cstdio> #include <queue> #include <cstring> using namespace std; inline int read(){ int s = 0, w = 1; char ch = getchar(); while(ch < '0' || ch > '9'){ if(ch == '-') w = -…