Attack Time Limit: 5000/3000 MS (Java/Others)    Memory Limit: 65768/65768 K (Java/Others)Total Submission(s): 2496    Accepted Submission(s): 788 Problem Description Today is the 10th Annual of “September 11 attacks”, the Al Qaeda is about to attack…

Coder Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 4838    Accepted Submission(s): 1853 Problem Description In mathematics and computer science, an algorithm describes a set of procedures…

Man Down Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 2030    Accepted Submission(s): 743 Problem Description The Game “Man Down 100 floors” is an famous and interesting game.You can enjoy t…

题目地址:pid=5326">HDU 5326 题意:给一张有向图n个点.n - 1(....输入n-1)条边. A指向B代表A管理B.然后能够间接管理,比方A管理B,B管理C.则A管理C. 如今问管理k个人的人有多少个. #include <stdio.h> #include <math.h> #include <string.h> #include <stdlib.h> #include <iostream> #include…

[HDU 2196] Computer(树的直径) 题链http://acm.hdu.edu.cn/showproblem.php?pid=2196 这题可以用树形DP解决,自然也可以用最直观的方法解,先求出树的直径,那么树的任意节点的最远点必然是直径上的两个端点之一,证明可以通过反证法构造: 设端点为a,b;设任意点为i,假设存在一点c到i的距离大鱼i到a,b的距离,那么a与c又能形成一个距离更长的点对,与ab是直径的假设不符,因此不存在c,证明完成. #include <queue> #i…

题意:给一棵树,每个结点的子树下的结点都是它的统治对象,问有多少个统治对象数目为k的结点? 思路:每个结点都设一个cnt来记数,只要将每个结点往上回溯,直到树根,经过的每个结点都将计数器加1.最后再扫一遍,将cnt等于k的结点统计一下. #include <bits/stdc++.h> #define LL long long #define pii pair<int,int> #define INF 0x7f7f7f7f using namespace std; ; int cn…

题目链接: http://acm.split.hdu.edu.cn/showproblem.php?pid=5861 Road Time Limit: 12000/6000 MS (Java/Others)Memory Limit: 65536/65536 K (Java/Others) 问题描述 There are n villages along a high way, and divided the high way into n-1 segments. Each segment woul…

Game Time Limit:1500MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Practice HDU 5242 Appoint description:  System Crawler  (2015-05-26) Description It is well known that Keima Katsuragi is The Capturing God because of his…

HDU 3333:http://acm.hdu.edu.cn/showproblem.php?pid=3333 这两个题是类似的,都是离线处理查询,对每次查询的区间的右端点进行排序.这里我们需要离散化处理一下,标记一下前面是否出现过这个值,然后不断更新last数组(该数组保存的是每个数最后一次出现的位置).最后用树状数组维护. #include<bits/stdc++.h> using namespace std; typedef long long ll; ; ; struct node{…

1.HDU 5877  Weak Pair 2.总结:有多种做法,这里写了dfs+线段树(或+树状树组),还可用主席树或平衡树,但还不会这两个 3.思路:利用dfs遍历子节点,同时对于每个子节点au,查询它有多少个祖先av满足av<=k/au. (1)dfs+线段树 #include<iostream> #include<cstring> #include<cmath> #include<queue> #include<algorithm>…