Watchcow Time Limit: 3000MS Memory Limit: 65536K Total Submissions: 9974 Accepted: 4307 Special Judge Description Bessie's been appointed the new watch-cow for the farm. Every night, it's her job to walk across the farm and make sure that no evildoer…

主题链接: http://poj.org/problem? id=2230 Watchcow Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 6055   Accepted: 2610   Special Judge Description Bessie's been appointed the new watch-cow for the farm. Every night, it's her job to walk ac…

http://poj.org/problem?id=3264 题意:给出n个数,还有q个询问,询问[l,r]区间里面最大值和最小值的差值. 思路:RMQ模板题,开两个数组维护最大值和最小值就行. #include <cstdio> #include <cstring> #include <cmath> #include <algorithm> using namespace std; #define N 50010 #define INF 0x3f3f3f3…

达神主席树讲解传送门:http://blog.csdn.net/dad3zz/article/details/50638026 2016-02-23:真的是模板题诶,主席树模板水过.今天新校网不好,没有评测,但我立下flag这个代码一定能A.我的同学在自习课上考语文,然而机房党都跑到机房来避难了\(^o^)/~ 2016-02-24:果然A乐~~~,我们机房党又躲过啦数学考试.良心不安啊 #include<cstdio> #include<cstring> #include<…

<题目链接> 题目大意: 给你一个H*W的矩阵,再告诉你有n个坐标有点,问你一个w*h的小矩阵最多能够包括多少个点. 解题分析:二维树状数组模板题. #include <cstdio> #include <cstring> #include <algorithm> using namespace std; +; int tr[M][M]; int n,W,H,w,h; int lowbit(int x){return x&(-x);} void ad…

题意 给定字符串A.B,求其最长公共子串 后缀数组模板题,求出height数组,判断sa[i]与sa[i-1]是否分属字符串A.B,统计答案即可. #include <cstdio> #include <cstdlib> #include <cstring> #include <string> #include <algorithm> #include <iostream> using namespace std; ; char st…

Bessie's been appointed the new watch-cow for the farm. Every night, it's her job to walk across the farm and make sure that no evildoers are doing any evil. She begins at the barn, makes her patrol, and then returns to the barn when she's done. If s…

Description Bessie's been appointed the new watch-cow for the farm. Every night, it's her job to walk across the farm and make sure that no evildoers are doing any evil. She begins at the barn, makes her patrol, and then returns to the barn when she'…

题目链接:http://poj.org/problem?id=2230 题目大意:给你n个点m条边,Bessie希望能走过每条边两次,且两次的方向相反,让你输出以点的形式输出路径. 解题思路:其实就是输出有向图的欧拉路,只是让你以点的形式输出.建图的时候,输入a,b直接建立(a,b)和(b,a)正反两条边,然后dfs递归遍历即可.注意输出点的位置:在边遍历完之后输出,原理暂时还没搞懂. 代码: #include<iostream> #include<cstdio> #include…

Watchcow Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 5258   Accepted: 2206   Special Judge Description Bessie's been appointed the new watch-cow for the farm. Every night, it's her job to walk across the farm and make sure that no ev…

关键是每条边必须走两遍,重复建边即可,因为确定了必然存在 Euler Circuit ,所以所有判断条件都不需要了. 注意:我是2500ms跑过的,鉴于这道题ac的code奇短,速度奇快,考虑解法应该不唯一. #include<cstdio> #include<cstring> #include<vector> #include<stack> #include<algorithm> #define rep(i,a,b) for(int i=a;i…

Watchcow Time Limit: 3000ms Memory Limit: 65536KB This problem will be judged on PKU. Original ID: 223064-bit integer IO format: %lld      Java class name: Main   Bessie's been appointed the new watch-cow for the farm. Every night, it's her job to wa…

Watchcow Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 8800   Accepted: 3832   Special Judge Description Bessie's been appointed the new watch-cow for the farm. Every night, it's her job to walk across the farm and make sure that no ev…

Watchcow Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 6336   Accepted: 2743   Special Judge Description Bessie's been appointed the new watch-cow for the farm. Every night, it's her job to walk across the farm and make sure that no ev…

http://poj.org/problem?id=3164 题意: 求最小树形图. 思路: 套模板. 引用一下来自大神博客的讲解:http://www.cnblogs.com/acjiumeng/p/7136604.html 算法步骤如下: 1.判断图的连通性,若不连通直接无解,否则一定有解. 2.为除了根节点以外的所有点选择一个权值最小的入边,假设用pre数组记录前驱,f数组记录选择的边长,记所选边权和为temp. 3.(可利用并查集)判断选择的的边是否构成环,若没有则直接$ans+=tem…

题目链接:http://poj.org/problem?id=2449 Time Limit: 4000MS Memory Limit: 65536K Description "Good man never makes girls wait or breaks an appointment!" said the mandarin duck father. Softly touching his little ducks' head, he told them a story. &quo…

http://poj.org/problem?id=2104 题意:求区间$[l,r]$的第k小. 思路:主席树不好理解啊,简单叙述一下吧. 主席树就是由多棵线段树组成的,对于数组$a[1,2...n]$,对于每个i,我们都去建立一棵线段树维护$a[1,..i]$出现的数的个数. 但是如果每一棵线段树都去新建结点的话,那这内存的开销是十分巨大的... 我们可以发现,第i棵线段树和第i-1棵线段树有很多结点都是相同的,这样一来,我们就没必要再去重新新建结点了,直接套用上一棵线段树的结点即可. 这里…

任意门:http://poj.org/problem?id=1330 Nearest Common Ancestors Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 34942   Accepted: 17695 Description A rooted tree is a well-known data structure in computer science and engineering. An example…

Kindergarten Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 6191   Accepted: 3052 Description In a kindergarten, there are a lot of kids. All girls of the kids know each other and all boys also know each other. In addition to that, some…

開始用瓜神说的方法撸了一发线段树.早上没事闲的看了一下树状数组的方法,于是又写了一发树状数组 树状数组: #include <cstdio> #include <cstring> #include <cmath> #include <algorithm> #include <climits> #include <string> #include <iostream> #include <map> #includ…

Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 44409   Accepted: 16184 Description A Bank plans to install a machine for cash withdrawal. The machine is able to deliver appropriate @ bills for a requested cash amount. The machine uses e…

题目链接:http://poj.org/problem?id=1273 Time Limit: 1000MS Memory Limit: 10000K Description Every time it rains on Farmer John's fields, a pond forms over Bessie's favorite clover patch. This means that the clover is covered by water for awhile and takes…

传送门:http://poj.org/problem?id=1655 题意:有T组数据,求出每组数据所构成的树的重心,输出这个树的重心的编号,并且输出重心删除后得到的最大子树的节点个数,如果个数相同,取编号小 的 思路:树的重心的模板题 首先要知道什么是树的重心,树的重心定义为:找到一个点,其所有的子树中最大的子树节点数最少,那么这个点就是这棵树的重心,删去重心后,生成的多棵树尽可能平衡.   所以寻找重心,即是最小化重心的最大子树.子树大小的计算分为两部分 的下面的子树大小,即可计算出 的上面…

POJ 1321-棋盘问题 K - DFS Time Limit:1000MS     Memory Limit:10000KB     64bit IO Format:%I64d & %I64u   Description 在一个给定形状的棋盘(形状可能是不规则的)上面摆放棋子,棋子没有区别.要求摆放时任意的两个棋子不能放在棋盘中的同一行或者同一列,请编程求解对于给定形状和大小的棋盘,摆放k个棋子的所有可行的摆放方案C. Input 输入含有多组测试数据. 每组数据的第一行是两个正整数,n k…

<题目链接> 题目大意: 给定一颗树,求出树的直径. 解题分析:树的直径模板题,以下程序分别用树形DP和两次BFS来求解. 树形DP: #include <cstdio> #include <algorithm> using namespace std; ; struct Edge{ int to,val,nxt; Edge(,,):to(_to),val(_val),nxt(_nxt){} }e[N<<]; int n,m,cnt,ans; int dp1…

<题目链接> 题目大意: 给出一棵树,问任意两个点的最近公共祖先的编号. 解题分析:LCA模板题,下面用的是树上倍增求解. #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> using namespace std; ; const int INF = 0x3f3f3f3f; struct Edge…

首先是当年stoer和wagner两位大佬发表的关于这个算法的论文:A Simple Min-Cut Algorithm 直接上算法部分: 分割线 begin 在这整篇论文中,我们假设一个普通无向图G=(V,E),其中每条边e都有一个正实数权值w(e). 如果我们知道:怎样找到两个节点s,t,以及怎样得到对于s-t的最小割,我们就几乎解决了整个问题: 定理2.1: 设s和t是图G中的两个节点,设G/{s,t}是合并s和t后得到的图, 则图G的全局最小割可以通过“图G对于s-t的最小割”和“图G/…

 题目链接 /* 模板题-------判断欧拉回路 欧拉路径,无向图 1判断是否为连通图, 2判断奇点的个数为0 */ #include <iostream> #include <cstring> #include <vector> #include <cstdio> using namespace std; struct DisjoinSet {//并查集判断是否连通 vector<int> father, rank; DisjoinSet(i…

http://poj.org/problem?id=3461 Oulipo Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 41051   Accepted: 16547 Description The French author Georges Perec (1936–1982) once wrote a book, La disparition, without the letter 'e'. He was a mem…

dfs算法模板: 1.下一层是多节点的dfs遍历 def dfs(array or root, cur_layer, path, result): if cur_layer == len(array) or not root: result.append(path) return for i in range(cur_layer, len(array)): do something with array[cur_layer:i+1] or nodes with this layer path.a…