TaoSama's Microcosm

HDU 2955 RobberiesTime Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64uSubmit StatusDescriptionThe aspiring Roy the Robber has seen a lot of American movie

题目链接： 轻点我一下 >3本题 类似于 HDU 1505 1506 不过本题列可移动 - - 机智的就是 可以通过辅助高度数组t 直接降序排序  从而达到t[i]*i即是最优子矩形面积的方便 然后扫一遍类似与1505 求出最大子矩形面积就好了AC代码如下：//// HDU 2830 Matrix Swapping II//// Created by TaoSama

HDU 1003  Max SumTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 158827    Accepted Submission(s): 37157Problem Description

HDU 1864 最大报销额Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 17530    Accepted Submission(s): 5141Problem Description现有一笔经费可以报销

HDU 1231 最大连续子序列Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 20624    Accepted Submission(s): 9144Problem Description给定K个

题目链接：轻点我一下~ >3求最大子矩形面积 = = 一开始很逗比的在求边界的时候写成了 - - l[i] = l[i]-1 r[i] = r[i] +1  n^2直接超时~ 正确的姿势应该是迭代 类似于动归的思想AC代码如下：//// HDU 1506 Largest Rectangle in a Histogram//// Created by TaoSama o

题目链接：轻点我一下 >3蛮有意思的一道题 开两个数组一个开灯on 一个关灯的off 然后分别对大小写字母进行决策就好了需要注意的是最后不要忘记关灯AC代码如下：//// HDU 2577 How to Type//// Created by TaoSama on 2015-02-08// Copyright (c) 2014 TaoSama. All rig

题目链接：轻点我一下 >3简单多重背包 - - 这个题不同于上一题的是 数据太大  需要二进制优化然后套01背包1s能跑10^8次 - - 运算不复杂的情况下 也是腻害的飞起AC代码如下：//// HDU 2844 Coins//// Created by TaoSama on 2015-02-08// Copyright (c) 2014 TaoSama.

FATETime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 8674    Accepted Submission(s): 4067Problem Description最近xhd正在玩一款叫做FATE的

HDU 2845 BeansTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3118    Accepted Submission(s): 1506Problem DescriptionBean-ea

Super Jumping! Jumping! Jumping!Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 24382    Accepted Submission(s): 10742Problem De

题目链接：轻点我一下 >3从下往上的姿势特别好 - - 其实可以优化下空间 直接把读入到dp数组中就可以了 懒得改了写的有点久了 然后状态方程变为dp[i][j] += max(dp[i + 1][j], dp[i + 1][j + 1])AC代码如下：//// HDU 2084 数塔//// Created by TaoSama on 2015-01-23// Co

题目链接： 轻点我一下 >3= = 看了题解的那个图 瞬间就明白了这题就是数塔 - - 还是从下往上的姿势 注意处理0的边界情况AC代码如下：//// HDU 1176 免费馅饼//// Created by TaoSama on 2015-02-07// Copyright (c) 2014 TaoSama. All rights reserved.//#i

题目链接： 轻点我一下 >3简单多重背包 需要注意的是背包大小是sum/2 输出的时候需要注意 左边的大于等于右边的写了O(nvk)的 - - 题目数据不是很大 就没写二进制优化 - - 其实是我忘记了QAQ半夜脑残  - - 50*50 = 250 果然我是250 数组少算一个0 RE了一发好吧 AC代码如下：//// HDU 1171 Big Event in

Bone CollectorTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 34170    Accepted Submission(s): 14054Problem DescriptionMany

命运Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 10875    Accepted Submission(s): 3837Problem Description穿过幽谷意味着离大魔王lemon已经

题目链接： 轻点我一下 >3本题和 HDU1506 很像 不过本题要对前i行求最大子矩形面积 进而可得最大子矩形面积 - - 需要注意的是关于高度的读入 要仔细想一想不要忘记每次清零h高度数组AC代码如下：//// HDU 1505 City Game//// Created by TaoSama on 2015-02-07// Copyright (c) 2

Largest SubmatrixTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1690    Accepted Submission(s): 805Problem DescriptionNow h

Monkey and BananaTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 8274    Accepted Submission(s): 4272Problem DescriptionA gr

SumsetsTime Limit: 2000MS Memory Limit: 200000KTotal Submissions: 13788 Accepted: 5480DescriptionFarmer John commanded his cows to search for different sets of

Cow BowlingTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 14627 Accepted: 9724DescriptionThe cows don't use actual bowling balls when they go bowling

Cheapest PalindromeTime Limit: 2000MS Memory Limit: 65536KTotal Submissions: 6243 Accepted: 3041DescriptionKeeping track of all the cows can be a tricky task so

Space ElevatorTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 8929 Accepted: 4232DescriptionThe cows are going to space! They plan to achieve orbit by

Bridging signalsTime Limit: 1000MS Memory Limit: 10000KTotal Submissions: 11270 Accepted: 6146Description'Oh no, they've done it again', cries the chief designe

Making the GradeTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 4404 Accepted: 2094DescriptionA straight dirt road connects two fields on FJ's farm, b

Cow ExhibitionTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 9687 Accepted: 3747Description"Fat and docile, big and dumb, they look so stupid, they a

Apple CatchingTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 8363 Accepted: 4093DescriptionIt is a little known fact that cows love apples. Farmer Jo

Milking TimeTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 5266 Accepted: 2172DescriptionBessie is such a hard-working cow. In fact, she is so focuse

Dollar DayzTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 4442 Accepted: 1716DescriptionFarmer John goes to Dollar Days at The Cow Store and discover

Ant CountingTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 3111 Accepted: 1233DescriptionBessie was poking around the ant hill one day watching the a

Wooden SticksTime Limit: 1000MS Memory Limit: 10000KTotal Submissions: 18903 Accepted: 7954DescriptionThere is a pile of n wooden sticks. The length and weight

CoinsTime Limit: 3000MS Memory Limit: 30000KTotal Submissions: 29974 Accepted: 10171DescriptionPeople in Silverland use coins.They have coins of value A1,A2,A3.

X-factor ChainsTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 5605 Accepted: 1770DescriptionGiven a positive integer X, an X-factor chain of length m

AB 时间太多久远 不记得了 - - 这次来补题 - - C和D 刚学了拓扑排序 迫不及待来补这个C 挺不错的题目C. Fox And Namestime limit per test2 secondsmemory limit per test256 megabytesinputstandard inp

Food Problem题意: 给你n种食物，m种车，每种食物有三种属性能量值t，体积u，数量v。每种车有三个属性值容量x,价格y,数量z。         问题是在能够达到至少p能量的要求下，最小花费为多少，若大于50000则输出TAT 分析: 两次多重背包dp 先dp出至少p能量的最小体积 然后从50000花费再dp出体积 在满足之前的最小体积下找到答案         为

题意: 用n分析: dp 貌似火柴的题 应该是这种数字的题 都有差不多的状态 首先将等式变成B+C=A 易于计算         考虑状态dp[left][carry][placeB][placeC] := dp[500][0/1][0/1][0/1]         从低位到位放置数字 剩余多少根火柴 是否有进位 B是否放置 C是否放置         至于怎么转移 看第一种情况

题意: 30*30的网格 从(1,1)走到(n,m) 每次只能向右或者向下 求路径的最小s2∗N2s^2*N^2 N为路径经过点数 分析: 都乘N2N^2了其实都告诉你 答案可以消掉除法了 然后化简s2s^2下得到 s2=x21+x22+...+x2nn−(x1+x2+...+xn)2n2s^2 = \frac {x_1^2 + x_2^2 + ... + x_n^2} {n} - \frac

A. Multiplication Table题意: 水题 - -//// Created by TaoSama on 2015-09-11// Copyright (c) 2015 TaoSama. All rights reserved.////#pragma comment(linker, "/STACK:1024000000,1024000000")#include

题意: 给定一个长度为N<=105N<=10^5的序列 求删去长度为0<=L<=N0<=L<=N 的序列后的LISLIS分析: 假设你已经会O(nlogn)O(nlogn)的LIS　删除连续一段长度为LL的序列 左边是i−L−1i - L - 1, 右边是ii, 那么去掉连续的mm个 求剩下的LISLIS 首先预处理出: f[i]:=f[i]:=以a[i]结尾的

题意: N<=106单词,给出每个单词的单词字母个数,并给出N个单词之间的修饰关系N<=10^6单词, 给出每个单词的单词字母个数, 并给出N个单词之间的修饰关系 任意两个修饰单词与被修饰单词的花费 是中间夹着的所有单词的长度和 求如何安排单词之间的顺序, 使得任意两个修饰单词与被修饰单词花费的总和最小, 输出这个总和分析: 对于多个修饰词同时修饰一个被修饰词的情况, 我们自然