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(HDU 1024)Max Sum Plus Plus dp
Max Sum Plus Plus Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 25894 Accepted Submission(s): 9004Problem Description Now I think you ha原创 2016-10-09 13:15:22 · 298 阅读 · 0 评论 -
(HDU 1074)Doing Homework dp 状态压缩
Doing Homework Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 8227 Accepted Submission(s): 3793Problem Description Ignatius has just come原创 2016-10-09 21:26:41 · 297 阅读 · 0 评论 -
(ZOJ 3469)Food Delivery 区间DP
Description When we are focusing on solving problems, we usually prefer to stay in front of computers rather than go out for lunch. At this time, we may call for food delivery.Suppose there are N peop原创 2016-10-21 13:32:39 · 492 阅读 · 0 评论 -
(HDU 1259) 最少拦截系统 dp || 贪心
最少拦截系统 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 35285 Accepted Submission(s): 13963Problem Description 某国为了防御敌国的导弹袭击,发展出一种导弹拦截系统.但是原创 2016-10-11 17:05:55 · 372 阅读 · 0 评论 -
(HDU 1160) FatMouse's Speed dp (最长递增子序列 , 两个因素决定)
FatMouse’s Speed Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 14760 Accepted Submission(s): 6502 Special JudgeProblem Description FatM原创 2016-10-11 23:35:18 · 313 阅读 · 0 评论 -
(POJ 1015) Jury Compromise 经典dp问题 (n选m)
转载请注明出处:優YoU http://blog.youkuaiyun.com/lyy289065406/article/details/6671105大致题意: 在遥远的国家佛罗布尼亚,嫌犯是否有罪,须由陪审团决定。陪审团是由法官从公众中挑选的。先随机挑选n 个人作为陪审团的候选人,然后再从这n 个人中选m 人组成陪审团。选m 人的办法是:控方和辩方会根据对候选人的喜欢程度,给所有候选人打分,分值从0转载 2016-10-12 16:32:49 · 417 阅读 · 0 评论 -
(POJ 1661) Help Jimmy 树型dp
Help Jimmy Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 10561 Accepted: 3464 Description“Help Jimmy” 是在下图所示的场景上完成的游戏。 场景中包括多个长度和高度各不相同的平台。地面是最低的平台,高度为零,长度无限。 Jimmy老鼠在时刻原创 2016-10-15 14:08:29 · 291 阅读 · 0 评论 -
(HDU 2859) Phalanx (求最大对称子矩阵) dp
Phalanx Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1246 Accepted Submission(s): 617Problem Description Today is army day, but the se原创 2016-10-15 15:11:23 · 1252 阅读 · 0 评论 -
(HDU 1078) FatMouse and Cheese 记忆化搜索
FatMouse and Cheese Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 8705 Accepted Submission(s): 3653Problem Description FatMouse has stor原创 2016-10-15 16:15:47 · 479 阅读 · 0 评论 -
POJ ACM训练指南
初期: 一.基本算法: (1)枚举. (poj1753,poj2965) (2)贪心(poj1328,poj2109,poj2586) (3)递归和分治法. (4)递推. (5)构造法.(poj3295) (6)模拟法.(poj1068,poj2632,poj1573,poj2993,poj2996) 二.图算法:转载 2016-09-30 13:33:49 · 478 阅读 · 0 评论 -
(POJ1035)Spell checker 字符串的处理
Spell checker Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 24755 Accepted: 9055 DescriptionYou, as a member of a development team for a new spell checking program, are to w原创 2016-09-30 20:05:47 · 487 阅读 · 0 评论 -
(POJ 3080)Blue Jeans 枚举 + KMP
Blue Jeans Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 16889 Accepted: 7493 DescriptionThe Genographic Project is a research partnership between IBM and The National Geogr原创 2016-09-30 20:13:18 · 665 阅读 · 0 评论 -
Nearest Neighbor Search 简单几何(求空间一点到区域的距离)
题目链接: https://acm.bnu.edu.cn/v3/statments/52296.pdf题意: 求空间一点(x0,y0,z0),到三维空间区域的最小距离?分析: 先求二维平面的最小距离(将平面分为9部分分别处理即可),再求高的距离。 AC代码:#include <iostream>#include <cstdio>#include <cstring>#include <a原创 2016-10-16 16:44:52 · 495 阅读 · 0 评论 -
Coins 思维题
题目链接: https://acm.bnu.edu.cn/v3/statments/52297.pdf题意: 分别有a1,a2,a3个一元,两元,3元的硬币。问可以组成多少不同的钱数?分析: 分类讨论即可,不过要用手推很长时间。 我们没用long long,很早就写出来了 ,WA了n次。。。。AC代码:#include <iostream>#include <cstdio>#includ原创 2016-10-16 17:24:44 · 528 阅读 · 0 评论 -
(POJ 1036)All in All 判断一个串是否包含另一个串 模拟
All in All Time Limit: 1000MS Memory Limit: 30000K Total Submissions: 32162 Accepted: 13387 DescriptionYou have devised a new encryption technique which encodes a message by inserting be原创 2016-09-30 20:56:36 · 418 阅读 · 0 评论 -
(POJ 2299)Ultra-QuickSort 归并排序求逆序数
Ultra-QuickSort Time Limit: 7000MS Memory Limit: 65536K Total Submissions: 56331 Accepted: 20814 DescriptionIn this problem, you have to analyze a particular sorting algorithm. The algor原创 2016-10-01 10:17:13 · 433 阅读 · 0 评论 -
(2016弱小联盟十一专场10.3)Parentheses 找规律
Parentheses Input: Standard Input Time Limit: See AtCoder Dave loves strings consisting only of (' and)’. Especially, he is interested in balanced strings. Any balanced strings can be constructed u原创 2016-10-26 22:44:39 · 685 阅读 · 0 评论 -
(2016弱小联盟十一专场10.3) Best Matched Pair 暴力
Best Matched Pair Input: Standard Input Time Limit: See AtCoder You are working for a worldwide game company as an engineer in Tokyo. This company holdsan annual event for all the staff members of t原创 2016-10-26 22:52:36 · 600 阅读 · 0 评论 -
(2016弱小联盟十一专场10.3) Help the Princess! BFS
题目连接: https://acm.bnu.edu.cn/v3/statments/jag2016.pdf分析:直接判断‘%’到‘@’和最近的‘$’的距离,如果 ‘%’到‘@’小于%’到最近的‘$’的距离则输出YES否则输出NOAC代码:#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#in原创 2016-10-26 22:59:53 · 438 阅读 · 0 评论 -
(POJ 3468)A Simple Problem with Integers 线段树区间修改入门讲解
C - A Simple Problem with Integers Time Limit:5000MS Memory Limit:131072KB 64bit IO Format:%lld & %lluPOJ 3468 Description You have N integers, A1, A2, … , AN. You need to deal with two kind原创 2016-10-02 17:13:47 · 434 阅读 · 0 评论 -
(HDU 1754)I Hate It 线段树区间查询入门,单点修改
I Hate It Time Limit:3000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u HDU 1754 Description 很多学校流行一种比较的习惯。老师们很喜欢询问,从某某到某某当中,分数最高的是多少。 这让很多学生很反感。 不管你喜不喜欢,现在需要你做的是,就是按照老师的要求,写一个程序,模原创 2016-10-02 17:17:35 · 415 阅读 · 0 评论 -
(POJ 2528)Mayor's posters 线段树 + 离散化
Mayor’s posters Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 59580 Accepted: 17267 DescriptionThe citizens of Bytetown, AB, could not stand that the candidates in the mayor原创 2016-10-03 13:43:11 · 312 阅读 · 0 评论 -
(HDU 1698)Just a Hook 线段树区间修改
Just a Hook Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 28817 Accepted Submission(s): 14276Problem Description In the game of DotA, Pu原创 2016-10-03 14:34:38 · 369 阅读 · 0 评论 -
(2016 弱校联盟十一专场10.3) Similarity of Subtrees DFS + hash
题目链接 https://acm.bnu.edu.cn/v3/problem_show.php?pid=52310problem description Define the depth of a node in a rooted tree by applying the following rules recursively: • The depth of a root node原创 2016-10-27 15:39:06 · 602 阅读 · 0 评论 -
Zoj 3537 Cake (DP_最优三角形剖分)
转自:http://blog.youkuaiyun.com/woshi250hua/article/details/7824433题目链接:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3537题目大意:给定n个点的坐标,先问这些点是否能组成一个凸包,如果是凸包,问用不相交的线来切这个凸包使得凸包只由三角形组成,根据costi, j =转载 2016-10-18 15:24:41 · 497 阅读 · 0 评论 -
(POJ 2955)Brackets 区间DP 最大括号匹配
Brackets Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 6647 Accepted: 3577 DescriptionWe give the following inductive definition of a “regular brackets” sequence:the empty sequ原创 2016-10-18 16:39:42 · 543 阅读 · 0 评论 -
(2016弱校联盟十一专场10.5) F. Fibonacci of Fibonacci (暴力 + 循环节)
The math class is coming again. You can’t wait to tell Nozomi the game “Knight Garden”, and she return a puzzle called “Fibonacci of Fibonacci” back to you. We are all familiar with Fibonacci sequenc转载 2016-10-28 14:36:40 · 715 阅读 · 0 评论 -
(POJ 3259)Wormholes 判断负环 bellman_ford 或者 spfa
Wormholes Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 45772 Accepted: 16904 DescriptionWhile exploring his many farms, Farmer John has discovered a number of amazing wormh原创 2016-10-28 16:54:42 · 492 阅读 · 0 评论 -
(POJ 1062)昂贵的聘礼 在可选的顶点中求最短路 dijkstra + 枚举
昂贵的聘礼 Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 46354 Accepted: 13753 Description年轻的探险家来到了一个印第安部落里。在那里他和酋长的女儿相爱了,于是便向酋长去求亲。酋长要他用10000个金币作为聘礼才答应把女儿嫁给他。探险家拿不出这么多金币,便请求酋长降低转载 2016-10-28 18:28:52 · 557 阅读 · 0 评论 -
(POJ 1651)Multiplication Puzzle 区间DP (dp思路总结)
Multiplication Puzzle Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 8806 Accepted: 5516 DescriptionThe multiplication puzzle is played with a row of cards, each containing a si原创 2016-10-19 18:49:59 · 490 阅读 · 0 评论 -
(LightOJ 1422 Halloween Costumes) 区间DP
Halloween Costumes Time Limit:2000MS Memory Limit:32768KB 64bit IO Format:%lld & %lluDescription Gappu has a very busy weekend ahead of him. Because, next weekend is Halloween, and he is plan原创 2016-10-19 18:55:55 · 378 阅读 · 0 评论 -
(POJ 2253)Frogger 求所有可达路径中的最大边的最小值 dijkstra || floyd 变形
Frogger Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 38822 Accepted: 12492 DescriptionFreddy Frog is sitting on a stone in the middle of a lake. Suddenly he notices Fiona F原创 2016-10-29 18:21:03 · 1218 阅读 · 0 评论 -
(POJ 2259) Team Queue 队列的应用 (队列的队列)
Team Queue Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 4403 Accepted: 1523 DescriptionQueues and Priority Queues are data structures which are known to most computer scientis原创 2016-10-29 18:30:15 · 617 阅读 · 0 评论 -
(2016年中国大学生程序设计竞赛(杭州)-重现赛) ArcSoft's Office Rearrangement 模拟
ArcSoft’s Office Rearrangement Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 0 Accepted Submission(s): 0Problem Description ArcSoft, Inc原创 2016-10-29 18:37:19 · 754 阅读 · 0 评论 -
(2016年中国大学生程序设计竞赛(杭州)-重现赛) Car 二分 (解决精度问题)
Car Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 0 Accepted Submission(s): 0Problem Description Ruins is driving a car to participating原创 2016-10-29 18:53:32 · 863 阅读 · 0 评论 -
(POJ2635)The Embarrassed Cryptographer <高精度求模(千进制表示) + 同余模定理 + 素数打表>
The Embarrassed Cryptographer DescriptionThe young and very promising cryptographer Odd Even has implemented the security module of a large system with thousands of users, which is now in use in his c原创 2016-11-29 13:37:27 · 559 阅读 · 0 评论 -
(POJ3292)Semi-prime H-numbers <素数筛法的变形>
Semi-prime H-numbers DescriptionThis problem is based on an exercise of David Hilbert, who pedagogically suggested that one study the theory of 4n+1 numbers. Here, we do only a bit of that.An H-number原创 2016-11-29 14:37:54 · 472 阅读 · 0 评论 -
(POJ 3041)Asteroids 二部图最大匹配 匈牙利算法
Asteroids Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 21027 Accepted: 11421 DescriptionBessie wants to navigate her spaceship through a dangerous asteroid field in the sha原创 2016-11-16 20:20:51 · 612 阅读 · 0 评论 -
(POJ1845)Sumdiv <约数和定理 + 快速幂求模 + 二分求等比数列和的模>
Sumdiv DescriptionConsider two natural numbers A and B. Let S be the sum of all natural divisors of A^B. Determine S modulo 9901 (the rest of the division of S by 9901). InputThe only line contains t原创 2016-11-29 19:10:32 · 835 阅读 · 0 评论 -
(POJ1905)Expanding Rods <几何+二分法(解方程)>
Expanding Rods DescriptionWhen a thin rod of length L is heated n degrees, it expands to a new length L’=(1+n*C)*L, where C is the coefficient of heat expansion. When a thin rod is mounted on two so原创 2016-11-30 14:00:37 · 438 阅读 · 0 评论