ACM-ICPC 2018 徐州赛区网络预赛 J. Maze Designer J. Maze Designer After the long vacation, the maze designer master has to do his job. A tour company gives him a map which is a rectangle. The map consists of N \times MN×M little squares. That is to say, the h…

https://nanti.jisuanke.com/t/31462 要求在一个矩形中任意选两个点都有唯一的通路,所以不会建多余的墙. 要求满足上述情况下,建墙的费用最小.理解题意后容易想到首先假设全部墙都建起来,然后拆掉费用最大的边使图成为一棵树,就是求一颗最大生成树 求出最大生成树后,求任意两点的距离,直接用lca就可以 思路 #include<bits/stdc++.h> #define M 300005 #define pb push_back using namespace std;…

大概就是要每两个点 只能有一条路径,并且约束,最短的边用来砌墙,那么反之的意思就是最大的边用来穿过 故最大生成树 生成以后 再用lca计算树上两点间的距离 (当然防止生成树是一条链,可以用树的重心作为根) ,然后简单的统计下树上两点的距离就行了 #include <bits/stdc++.h> using namespace std; **; int n, m, sum, fa[N]; struct node { int u,v,val; node(){} node(int uu,int vv…

https://nanti.jisuanke.com/t/31462 题意 一个N*M的矩形,每个格点到其邻近点的边有其权值,需要构建出一个迷宫,使得构建迷宫的边权之和最小,之后Q次查询,每次给出两点坐标,给出两点之间的最短路径 分析 可以把每个格点视作视作图的点,隔开两点的边视作图的边,则构建迷宫可以视作求其生成树,剩余的边就是组成迷宫的墙.因为要花费最小,所以使删去的墙权置最大即可,呢么就是求最大生成树即可.然后每次查询相当于查这个最大生成树上任意两点的最短距离,到这一步就是个LCA了. 这…

传送门:https://nanti.jisuanke.com/t/31462 本题是一个树上的问题:结点间路径问题. 给定一个有N×M个结点的网格,并给出结点间建立墙(即拆除边)的代价.花费最小的代价,使得每一对结点之间的路径唯一.给出Q次询问:每次询问一对结点的路径长度. 每一对结点之间存在路径,则图是连通的:路径唯一,则图是无环的.于是拆除边后的图是原图的一棵生成树.为使得拆除的代价尽可能小,这棵生成树应是最大生成树.通过Kruskal算法,可以求解最大生成树. 之后,即是询问树结点对的路径…

ACM-ICPC 2018 徐州赛区网络预赛 G. Trace (思维,贪心) Trace 问答问题反馈 只看题面 35.78% 1000ms 262144K There's a beach in the first quadrant. And from time to time, there are sea waves. A wave ( xx , yy ) means the wave is a rectangle whose vertexes are ( 00 , 00 ), ( xx ,…

H.Ryuji doesn't want to study 27.34% 1000ms 262144K   Ryuji is not a good student, and he doesn't want to study. But there are n books he should learn, each book has its knowledge a[i]a[i]. Unfortunately, the longer he learns, the fewer he gets. That…

传送门 题面: In a world where ordinary people cannot reach, a boy named "Koutarou" and a girl named "Sena" are playing a video game. The game system of this video game is quite unique: in the process of playing this game, you need to consta…

In a world where ordinary people cannot reach, a boy named "Koutarou" and a girl named "Sena" are playing a video game. The game system of this video game is quite unique: in the process of playing this game, you need to constantly fac…

262144K   Morgana is learning computer vision, and he likes cats, too. One day he wants to find the cat movement from a cat video. To do this, he extracts cat features in each frame. A cat feature is a two-dimension vector <x, y>. If xi​= xj​ and yi…