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POJ 2000 Gold Coins
DescriptionThe king pays his loyal knight in gold coins. On the first day of his service, the knight receives one gold coin. On each of the next two days (the second and third days of service), the原创 2015-09-13 19:20:10 · 425 阅读 · 0 评论 -
poj 2524 Ubiquitous Religions(并查集)
Ubiquitous ReligionsTime Limit: 5000MS Memory Limit: 65536KTotal Submissions: 32572 Accepted: 15798DescriptionThere are so many different religions in the world原创 2016-11-11 17:12:28 · 276 阅读 · 0 评论 -
poj 1013 Counterfeit Dollar(模拟题)
Counterfeit DollarTime Limit: 1000MS Memory Limit: 10000KTotal Submissions: 45332 Accepted: 14324DescriptionSally Jones has a dozen Voyageur silver dollars. How原创 2016-11-12 17:06:26 · 647 阅读 · 0 评论 -
poj 1316 Self Numbers(水)
Self NumbersTime Limit: 1000MS Memory Limit: 10000KTotal Submissions: 23405 Accepted: 13142DescriptionIn 1949 the Indian mathematician D.R. Kaprekar discovered原创 2016-11-12 17:15:44 · 514 阅读 · 0 评论 -
poj 2251 Dungeon Master (BFS)
Dungeon MasterTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 28186 Accepted: 11006DescriptionYou are trapped in a 3D dungeon and need to find the qui原创 2016-11-12 17:33:49 · 301 阅读 · 0 评论 -
poj 1942 Paths on a Grid(组合数学--组合数)
Paths on a GridTime Limit: 1000MS Memory Limit: 30000KTotal Submissions: 25146 Accepted: 6274DescriptionImagine you are attending your math lesson at school. On原创 2016-11-12 21:48:41 · 560 阅读 · 0 评论 -
poj 3050 Hopscotch(dfs暴力)
HopscotchTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 3951 Accepted: 2627DescriptionThe cows play the child's game of hopscotch in a non-traditional原创 2017-02-14 10:59:00 · 411 阅读 · 0 评论 -
poj 3984 迷宫问题
迷宫问题Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 18533 Accepted: 10970Description定义一个二维数组: int maze[5][5] = { 0, 1, 0, 0, 0, 0, 1, 0, 1, 0,原创 2017-02-01 00:14:48 · 405 阅读 · 0 评论 -
poj 1008 Maya Calendar
Maya CalendarTime Limit: 1000MS Memory Limit: 10000KTotal Submissions: 76933 Accepted: 23686DescriptionDuring his last sabbatical, professor M. A. Ya made a sur原创 2017-02-01 00:20:48 · 489 阅读 · 0 评论 -
poj 2240 Arbitrage(Floyd最短路)
Arbitrage is the use of discrepancies in currency exchange rates to transform one unit of a currency into more than one unit of the same currency. For example, suppose that 1 US Dollar buys 0.5 British原创 2017-03-05 15:40:01 · 494 阅读 · 0 评论 -
poj 3414 Pots(BSF + 记录路径)
PotsTime Limit: 1000MS Memory Limit: 65536K Total Submissions: 15972 Accepted: 6730 Special JudgeDescriptionYou are given two pots, having the volume of A and B liters respectively. T原创 2017-03-10 20:39:47 · 424 阅读 · 0 评论 -
第二周---最小生成树(Kruskal,Prim)、拓扑排序
训练链接:https://vjudge.net/contest/159297#rankpoj 2403题意:给出职位名称,该职位的工资,然后给出一个工作的描述,要求出能给出多少工资。分析:字符串模拟,判断一段话中有没有该职位名称,有的话就把工资加上。AC代码:#include <stdio.h>#include <map>#include <iostream>using namespace s原创 2017-04-24 20:37:14 · 388 阅读 · 0 评论 -
poj 2084 Game of Connections (卡特兰数、大数)
Game of ConnectionsTime Limit: 1000MS Memory Limit: 30000KTotal Submissions: 8303 Accepted: 4133DescriptionThis is a small but ancient game. You are supposed to原创 2016-10-25 11:35:27 · 1835 阅读 · 0 评论 -
poj 2488 A Knight's Journey (dfs)
A Knight's JourneyTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 42114 Accepted: 14324DescriptionBackground The knight is getting bored of seeing t原创 2016-10-16 21:52:22 · 303 阅读 · 0 评论 -
poj 1573 Robot Motion (模拟题)
Robot MotionTime Limit: 1000MS Memory Limit: 10000KTotal Submissions: 13100 Accepted: 6349DescriptionA robot has been programmed to follow the instructions原创 2016-10-16 13:00:52 · 450 阅读 · 0 评论 -
POJ 3096 Surprising Strings(STL map)
Surprising StringsTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 6569 Accepted: 4268DescriptionThe D-pairs of a string of letters are the ordered pai原创 2016-04-26 09:10:45 · 527 阅读 · 0 评论 -
POJ 2965 The Pilots Brothers' refrigerator(回溯查找,dfs)
The Pilots Brothers' refrigeratorTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 23414 Accepted: 9004 Special JudgeDescriptionThe game “The Pilo原创 2016-06-25 20:10:27 · 321 阅读 · 0 评论 -
POJ 1338 Ugly Numbers(丑数)
Ugly NumbersTime Limit: 1000MS Memory Limit: 10000KTotal Submissions: 22726 Accepted: 10127DescriptionUgly numbers are numbers whose only prime factors are 2, 3原创 2016-06-29 15:34:22 · 550 阅读 · 0 评论 -
一切从头再来,poj分类
从今天起,一切题目从头刷起,从POJ开始,希望能上榜!POJ分类: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原创 2016-06-21 20:42:24 · 601 阅读 · 0 评论 -
POJ 1753 Flip Game(回溯查找,dfs)
Flip GameTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 39171 Accepted: 17026DescriptionFlip game is played on a rectangular 4x4 field with two-sided p原创 2016-06-23 16:06:48 · 438 阅读 · 0 评论 -
POJ 3050 Hopscotch (回溯枚举)
HopscotchTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 3373 Accepted: 2326DescriptionThe cows play the child's game of hopscotch in a non-traditiona原创 2016-09-10 20:19:14 · 381 阅读 · 0 评论 -
poj 1979 Red and Black(BFS)
Red and BlackTime Limit: 1000MS Memory Limit: 30000KTotal Submissions: 31221 Accepted: 17047DescriptionThere is a rectangular room, covered with square tiles. E原创 2016-09-04 11:16:05 · 655 阅读 · 0 评论 -
poj 1166 The Clocks (暴力|| dfs)
The ClocksTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 16196 Accepted: 6627Description|-------| |-------| |-------|| | | |原创 2016-09-20 13:38:16 · 490 阅读 · 0 评论 -
poj 3009 Curling 2.0
Curling 2.0Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 18688 Accepted: 7666DescriptionOn Planet MM-21, after their Olympic games this year, curlin原创 2016-10-22 11:52:01 · 466 阅读 · 0 评论 -
poj 1068 (Parencodings)
ParencodingsTime Limit: 1000MS Memory Limit: 10000KB 64bit IO Format: %lld & %lluSubmit StatusDescriptionLet S = s1 s2...s2n be a well-formed string of parenthe原创 2016-10-15 10:34:30 · 477 阅读 · 0 评论 -
poj 2632 Crashing Robots (模拟题)
Crashing RobotsTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 10286 Accepted: 4362DescriptionIn a modernized warehouse, robots are used to fetch the原创 2016-10-16 11:16:43 · 391 阅读 · 0 评论 -
第一周---最短路(Dijkstra、SPFA)
训练链接:https://vjudge.net/contest/157742Poj 2387题意:给出N个谷仓,T条路径,求出从1谷仓到N号谷仓的最短距离。分析:Dijkstra模板题目,只是需要注意判断重边的情况。AC代码:用邻接矩阵存储#include <stdio.h>#include <string.h>#define INF 1<<30int map[1005][1005原创 2017-04-17 16:37:15 · 384 阅读 · 0 评论