ACdream 1415 Important Roads

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ACdream 1415 Important Roads

Important Roads Special JudgeTime Limit: 20000/10000MS (Java/Others)Memory Limit: 128000/64000KB (Java/Others) Problem Description The city where Georgie lives has n junctions some of which are connected by bidirectional roads. Every day Geo…

Codeforces Gym 100338C C - Important Roads tarjan

C - Important RoadsTime Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://acm.hust.edu.cn/vjudge/contest/view.action?cid=88926#problem/C Description The city where Georgie lives has n junctions some of which are connected by bidirectional roads. Every d…

Codeforces Gym 100338C Important Roads 最短路+Tarjan找桥

原题链接:http://codeforces.com/gym/100338/attachments/download/2136/20062007-winter-petrozavodsk-camp-andrew-stankevich-contest-22-asc-22-en.pdf 题意给你一个无向图,要从1走到n,问你哪些边去掉之后就没法走原本的最短路了. 题解跑两发最短路,顺着跑一发,倒着跑一发,对于边(u,v),如果w(u,v)+d[u]+rd[v]或者w(u,v)+d[v]+rd[u]…

codeforces Gym 100338C Important Roads （重建最短路图）

正反两次最短路用于判断边是不是最短路上的边,把最短路径上的边取出来建图.然后求割边.注意重边,和卡spfa. 正权,好好的dijkstra不用,用什么spfa? #include<bits/stdc++.h> using namespace std; typedef long long ll; #define fi first #define se second #define bug(x) cout<<#x<<'='<<x<<endl; #de…

Gym - 100338C Important Roads 最短路+tarjan

题意:给你一幅图,问有多少条路径使得去掉该条路后最短路发生变化. 思路:先起始两点求两遍单源最短路,利用s[u] + t[v] + G[u][v] = dis 找出所有最短路径,构造新图.在新图中找到所有的桥输出就可以了. #include <iostream> #include <cstdio> #include <fstream> #include <algorithm> #include <cmath> #include <deque…

acdream1415(dij+优先队列+桥)

这题好坑,卡SPFA... 无奈只能用dij+优先队列了. 因为好久没有写过代码了,所以今天写dij时候突然觉得复杂度不对,dij+优先队列的复杂度是(n+m)logn,这种复杂度对于稠密图是非常慢!,而且还有超内存的可能(最坏情况要把n*n个点都存进优先队列),与我以前记得复杂度是nlogn不一样... 瞬间吓尿. 其实事实确实是这样,在采用斐波那契堆+dij时,斐波那契堆是插入复杂度为O(1),所以复杂度为nlogn+m,而普通我们用的STL种的priority_queue插入查询复杂度都是…