一開始想,总感觉是DP,但是最后什么都没想到.还暴力的交了一发. 然后開始写线段树,结果超时.感觉自己线段树的写法有问题.改天再写.先把树状数组的写法贴出来吧. ~~~~~~~~~~~~~~~~~~~~~~~~ 树状数组不懂的去看刘汝佳的大白书,那个图画得非常清楚. 题目大意:星星的坐标以y递增的顺序给出,这些点的左下方的点数代表这个点的级数,问0~N-1的级数有多少个?事实上y根本木实用. 题目链接:http://poj.org/problem?id=2352 http://acm.hdu.e…

题目http://acm.hdu.edu.cn/showproblem.php?pid=1541 n个星星的坐标,问在某个点左边(横坐标和纵坐标不大于该点)的点的个数有多少个,输出n行,每行有一个数字,代表左下角个点数为i( i 从0开始)的点的个数. 首先将横坐标排序,这样就值需要比较纵坐标,可以采用hash的方法记录纵坐标为 i 的个数,每枚举到一个点hash加一,然后树状数组求纵坐标 不大于该点的 个数.值得注意的是树状数组是由 1 开始的,而题目可以从0开始,会超时,加一就好. 具体看代…

Stars Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 13280    Accepted Submission(s): 5184 Problem Description Astronomers often examine star maps where stars are represented by points on a pl…

点我看题目 题意 :N个村子连成一条线,相邻的村子都有直接的地道进行相连,不相连的都由地道间接相连,三个命令,D x,表示x村庄被摧毁,R  ,表示最后被摧毁的村庄已经重建了,Q x表示,与x直接或间接相连的村庄有多少个,当然包括他自己. 思路 :这是一道用线段树,树状数组,还有STL都可以做的题....因为用线段树的更新什么的太麻烦,.....所以我用了树状数组 #include <iostream> #include <stdio.h> #include <string.…

http://acm.hdu.edu.cn/showproblem.php?pid=1541 Stars Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 5993    Accepted Submission(s): 2384 Problem Description Astronomers often examine star maps…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1541 题目大意就是统计其左上位置的星星的个数 由于y已经按升序排列,因此只用按照x坐标生成一维树状数组即可 比较简单的题目: 代码: #include<iostream> #include<cstring> #include<cstdio> #include<cstdlib> #include<algorithm> using namespace s…

Mobile phones Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 14489   Accepted: 6735 Description Suppose that the fourth generation mobile phone base stations in the Tampere area operate as follows. The area is divided into squares. The…

Reference: http://blog.csdn.net/me4546/article/details/6333225 题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=2838 题目大意:每头牛有个愤怒值,每次交换相邻两个数进行升序排序,$cost=val_{1}+val_{2}$,求$\min \sum cost_{i}$ 解题思路: 按输入顺序DP: 第i的值val的最小cost=当前数的逆序数个数*val+当前数的逆序数和 相当于每次只…

Sort it Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4110    Accepted Submission(s): 2920 Problem Description You want to processe a sequence of n distinct integers by swapping two adjacent s…

Stars Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 39186 Accepted: 17027 Description Astronomers often examine star maps where stars are represented by points on a plane and each star has Cartesian coordinates. Let the level of a star b…

Stars Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 37323 Accepted: 16278 Description Astronomers often examine star maps where stars are represented by points on a plane and each star has Cartesian coordinates. Let the level of a star b…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4046 When I wrote down this letter, you may have been on the airplane to U.S. We have known for 15 years, which has exceeded one-fifth of my whole life. I still remember the first time we went to the mov…

http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=10486 题意:一颗有n个分支的苹果树,根为1,每个分支只有一个苹果,给出n-1个分支的关系和给出m个操作,Q x表示询问x的子树(包括x)苹果的数量,C x表示若分支x上有苹果,则摘下来,若没有则会生出一个,输出每个询问的值. DFS序 每个子树对应的孩子节点包括根用dfs序存在连续区间中,用树状数组维护区间 //#pragma comment(linker, "/STAC…

Inversion Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=5497 Description 你有一个序列\{a_1,a_2,...,a_n\}{a​1​​,a​2​​,...,a​n​​},然后你可以删除一个长度为mm的连续子序列. 问如何删除才能使逆序对最少. Input 输入有多组数据, 第一行有一个整数TT表示测试数据的组数. 对于每组数据: 第一行包含2个…

Queue Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=5493 Description N people numbered from 1 to N are waiting in a bank for service. They all stand in a queue, but the queue never moves. It is lunch time now,…

题目链接:http://poj.org/problem?id=2533 其实这个题的数据范围n^2都可以过,只是为了练习一下nlogn的写法. 最长上升子序列的nlogn写法有两种,一种是变形的dp,另一种是树状数组. 变形的dp可以参考http://www.cnblogs.com/itlqs/p/5743114.html 树状数组的写法其实就是用到了树状数组求前缀最值,必要的时候可以离散化一下. #include<cstdio> #include<cstring> #includ…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4031 Attack Time Limit: 5000/3000 MS (Java/Others)    Memory Limit: 65768/65768 K (Java/Others) Total Submission(s): 1890    Accepted Submission(s): 554 Problem Description Today is the 10th Annual of "S…

Astronomers often examine star maps where stars are represented by points on a plane and each star has Cartesian coordinates. Let the level of a star be an amount of the stars that are not higher and not to the right of the given star. Astronomers wa…

题目链接:https://cn.vjudge.net/problem/HDU-1541 题意 天上有许多星星 现给天空一个平面坐标轴,统计每个星星的level, level是指某一颗星星的左下角(x<=x0 && y<=y0)的星星总数 注意数据以yx排序输入 思路 怎么想也想不到如何查询 若用传统O(n^3)查询必然超时 于是想到记忆化搜索不断读写dp[y][x]即可,目测O(n^2)大一点 算了一下时间空间都不大可行,苦思冥想没有思路,看了别人的题解才晃然大雾 用树状数组维…

Flowers Problem Description As is known to all, the blooming time and duration varies between different kinds of flowers. Now there is a garden planted full of flowers. The gardener wants to know how many flowers will bloom in the garden in a specifi…

题目:这里 题意: 给出一个n个结点的树和一个数k,每个结点都有一个权值,问有多少对点(u,v)满足u是v的祖先结点且二者的权值之积小于等于k. 从根结点开始dfs,假设搜的的点的权值是v,我们需要的是在此之前搜的点中小于等于k/v的数的个数,于是用树状数组查询,查完后将v加入树状数组以供下个 点查询,回溯的时候再一个个的删除将其树状数组.由于这个权值和k的范围比较大,所以得先离散化后在加入树状数组. #include<cstdio> #include<cstring> #incl…

传送门 The k-th Largest Group Time Limit: 2000MS   Memory Limit: 131072K Total Submissions: 8690   Accepted: 2847 Description Newman likes playing with cats. He possesses lots of cats in his home. Because the number of cats is really huge, Newman wants…

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2048    Accepted Submission(s): 805 Problem Description Recently, dobby is addicted in the Fruit Ninja. As you know, dobby is a free elf, so unlik…

                                                              Mobile phones Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 18608   Accepted: 8621 Description Suppose that the fourth generation mobile phone base stations in the Tampere a…

实际上就是问这个区间编号连续的段的个数,假如一个编号连续的段有(a+b)个人,我把他们分在同一组能得到的分值为(a+b)^2,而把他们分成人数为a和b的两组的话,得到的分值就是a^2+b^2,显然(a+b)^2 > a^2+b^2,所以对于每个区间,尽可能把他们分成尽量少的组. 考虑把这些数从后往前添加,每添加一个数num[i],如果num[i]+1或者num[i]-1有且只有一个已经存在,则段数不变:如果num[i]+1和num[i]-1同时存在,则段数-1,如果都不在,则段数+1. 对于每个…

题意 询问一段区间里的数能组成多少段连续的数. 思路 先考虑从左往右一个数一个数添加,考虑当前添加了i - 1个数的答案是x,那么可以看出添加完i个数后的答案是根据a[i]-1和a[i]+1是否已经添加而定的:如果a[i]-1或者a[i]+1已经添加一个,则段数不变,如果都没添加则段数加1,如果都添加了则段数减1.设v[i]为加入第i个数后的改变量,那么加到第x数时的段数就是sum{v[i]} (1<=i<=x}.当然,若删除某个数,那么这个数两端的数的改变量也会跟着改变,这样一段区间的数构成…

树状数组这个真心想了好久,还是没想出来 %%% www.cppblog.com/Yuan/archive/2010/08/18/123871.html 树状数组求前缀和大于等于k的最大值,第一次看到这种方法,很神奇,就是没看懂= = 二分也是可以求的,不过感觉会慢一些…… 思路就是把所有没有询问到的数压缩 例如如果n等于10 值询问到了 2, 7 大概是这样的 [1,2][3,4,5,6,7][8,9,10] 1                2                         …

Cow Patterns Description A particular subgroup of K (1 <= K <= 25,000) of Farmer John's cows likes to make trouble. When placed in a line, these troublemakers stand together in a particular order. In order to locate these troublemakers, FJ has lined…

偶然发现这题还没A掉............速速解决了............. 树状数组和线段树比较下,线段树是在是太冗余了,以后能用树状数组还是尽量用......... #include <iostream> #include <algorithm> #include <cmath> #include <cstdio> #include <cstdlib> #include <cstring> #include <strin…

三维树状数组模版.优化不动了. #include <set> #include <map> #include <stack> #include <cmath> #include <queue> #include <cstdio> #include <string> #include <vector> #include <iomanip> #include <cstring> #inclu…