Problem H Morning Walk Time Limit 3 Seconds Kamal is a Motashota guy. He has got a new job in Chittagong. So, he has moved to Chittagong from Dinajpur. He was getting fatter in Dinajpur as he had no work in his hand there. So, moving to Chittagong ha…

题目给出图,要求判断不能一遍走完所有边,也就是无向图,题目分类是分欧拉回路,但其实只要判断度数就行了. 一开始以为只要判断度数就可以了,交了一发WA了.听别人说要先判断是否是联通图,于是用并查集并一起,然后判断是否有多个根. 用dfs的话就是深搜时标记下,最后看看有没有全部标记.我没用dfs做. 代码: #include <cstdio> const int maxn = 201; int f[maxn]; int d[maxn]; int find(int x) { if (x != f[x…

题意:给出n个点,m条路,问能否走完m条路. 自己做的时候= =三下两下用并查集做了交,WA了一发-后来又WA了好几发--(而且也是判断了连通性的啊) 搜了题解= = 发现是这样的: 因为只要求走完所有的路,即为只需要走完已经给出的路,而并没有要求所走得路上含有所有的点, 比如说 给出的路有这些 0 1 1 2 2 3 3 0 4 4 那么构成的路即为,绕着图中的蓝色线走一圈,即为走完了所有的路, 而4是一个孤立点,也并没有构成路,所以不需要管它 代码中的 if(d[i]!=0)是判断这个点是否…

A Walk Through the Forest Time Limit:1000MS  Memory Limit:65536K Total Submit:48 Accepted:15 Description Jimmy experiences a lot of stress at work these days, especially since his accident made working difficult. To relax after a hard day, he likes t…

Problem H Morning Walk Time Limit 3 Seconds Kamalis a Motashotaguy. He has got a new job in Chittagong. So, he has moved to Chittagong fromDinajpur. He was getting fatter in Dinajpur as he had no work in his hand there. So, moving toChittagong has tu…

https://vjudge.net/problem/UVA-503 题目 给出一个长方体和长方体上两点的坐标,求两点的沿着长方体表面走的最小距离 题解 沿着表面走就是在展开图上面走,如果分类讨论就需要考虑很多情况,比如两个相邻的面.相对的面,有时候需要走4个展开面,有时候要走3个,是不是走的面越多距离越长,这些都说不清楚……而且手动写出所有情况很麻烦…… 于是只有选择模拟展开这个长方体了,需要考虑很多细节,比如给面编号,把每个点对应到面的坐标找出来,然后还要判断走展开图是否不会超出每个面 这时…

题目 Volume 0. Getting Started 开始10055 - Hashmat the Brave Warrior 10071 - Back to High School Physics 10300 - Ecological Premium 458 - The Decoder 494 - Kindergarten Counting Game 414 - Machined Surfaces 490 - Rotating Sentences 445 - Marvelous Mazes…

Problem    UVA - 10917 - Walk Through the Forest Time Limit: 3000 mSec Problem Description Jimmy experiences a lot of stress at work these days, especially since his accident made working difficult. To relax after a hard day, he likes to walk home. To…

uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1858 题目大意: Jimmy下班后决定每天沿着一条不同的路径回家,欣赏不同的风景.他打算只沿着满足如下条件的(A,B)道路走:存在一条从B出发回家的路,比所有从A出发回家的路径都短.你的任务是计算一共有多少条不同的回家路径.其中公司的编号为1,家的编号为2. 思路: 题目给出的n于1000以内,所以我直接…

用新模板阿姨了一天,换成原来的一遍就ac了= = 题意很重要..最关键的一句话是说:若走A->B这条边,必然是d[B]<d[A],d[]数组保存的是各点到终点的最短路. 所以先做dij,由d[B]<d[A]可知,所走的路径上各点的d[]值是由大到小的,即是一个DAG,从而决定用记忆化搜索查找总的路径数. #include<stdio.h> #include<string.h> #include<algorithm> using namespace st…