Codeforces - Pashmak and Graph 题目链接:Codeforces - Pashmak and Graph 我比较喜欢暴力,所以采用了暴力解法。 显然可以先对边排序,然后直接dp,分别记录是否相等权值?不,我喜欢暴力。 所以我们开N颗权值线段树暴力转移即可。 AC代码:...Codeforces - Greg and Graph 题目链接:Codeforces - Greg and Graph 考虑...
Codeforces 295B 《Greg and Graph》 原创建时间:2018-09-30 20:21:16 开倒车 倒序Floyd 题目链接题目描述翻译来自洛谷 Greg有一个有边权的有向图,包含 nn 个点。这个图的每两个点之间都有两个方向的边。Greg喜欢用他的图玩游戏,现在他发明了一种新游戏:游戏...
CodeForces 295B Greg and Graph 题解 \(Floyd\) 算法是一种基于动态规划的算法,以此题为例介绍最短路算法中的 \(Floyd\) 算法。 我们考虑给定一个图,要找出 \(i\) 号点到 \(j\) 号点的最短路径。则该最短路径只有两种可能: \(i\) 号点直接
Greg has a weighed directed graph, consisting ofnvertices. In this graph any pair of distinct vertices has an edge between them in both directions. Greg loves playing with the graph and now he has invented a new game: The game consists ofnsteps. On thei-th step Greg removes vertex number...
B. Greg and Graph time limit per test 3 seconds memory limit per test 256 megabytes input standard i
On thei-th step Greg removes vertex numberxifrom the graph. As Greg removes a vertex, he also removes all the edges that go in and out of this vertex. Before executing each step, Greg wants to know the sum of lengths of the shortest paths between all pairs of the remaining vertices....
CodeForces 295B Greg and Graph (floyd+离线) 题目链接> 题目大意: 给定$n$个点的有向完全带权图$(n\leq500)$,现在进行$n$次操作,每次操作从图中删除一个点(每删除一个点,都会将与它相关联的边都删除),问你每次删点之前,图中所有点对的最近距离之和。
Greg has a weighed directed graph, consisting of n vertices. In this graph any pair of distinct vertices has an edge between them in both directions. Greg loves playing with the graph and now he has invented a new game: The game consists of n steps. On the i-th step Greg removes ve...
CodeForces 295B Greg and Graph (floyd+离线) <题目链接> 题目大意:给定nn个点的有向完全带权图(n≤500)(n≤500),现在进行nn次操作,每次操作从图中删除一个点(每删除一个点,都会将与它相关联的边都删除),问你每次删点之前,图中所有点对的最近距离之和。 解题分析: 删除操作不好实现,逆向思维,从后...
设新加入的点为x,则我们枚举u,v(都是从1到n),然后更新 a[u][v] = min(a[u][v], a[u][x] + a[x][v]); 这样一来只要计算答案时,枚举的u,v都是当前已经加入的点,就能保证计算的所有最短路经过的点都是当前存在的点。 #include<iostream>#include<cstdio>usingnamespacestd;#definell long lo...