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RSS订阅原创 UVa 1509 Leet 枚举
题意：有两条字符串alph和leet，alph由纯字母组成，长度小于等于15，能够映射到leet，每个字母能对应k个leet字符，对应后不再改变。1≤k≤3如果alph中所有字符映射成功，输出1；否则输出0。分析：数据量小，可以枚举解决。alph是纯字母，所以可以放在大小为26的数组里，每个元素存储k个对应字符。代码：#include #include #include
2016-05-30 14:06:49
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原创 UVa 1629 Cake Slicing DP
这题空了好久，虽然现在想出来了很激动，但是花的时间略长呀。。。每切一次蛋糕就相当于一次状态转移。设某时刻蛋糕的坐上顶点为（a, b），右下顶点为（c， d），将这块蛋糕切成只剩一个樱桃产生的切割线长度最短为d （ a， b， c， d）。原来是想根据樱桃的位置来切蛋糕，但是很难实现。考虑到横竖共m + n 种切法，可以枚举状态。纵向：min { d ( a, b, c, j ) + d
2015-12-15 05:50:18
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原创 poj 1971 Parallelogram Counting 排序 + 计数
题意：给出平面上的n个点，求出组成的平行四边形个数。考虑平行四边形的性质，对角线互相平分。因为做之前提示可以哈希，所以想把所有中点哈希后计数，但是这样太麻烦。直接对所有点排序后，将重复的计数t，求所有t * (t - 1) / 2的和。#include #include #include #include using namespace std;const int
2015-10-14 08:58:01
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原创 poj 2002 Squares 哈希
题意：给出平面上的n个点，求出能组成的正方形个数。（n 先考虑暴力，O（n^4），会超时。再考虑正方形的特点，已知两个点，可以推出另外两个点。所以枚举两个点，因为点的坐标|x，y|枚举两个点时，我曾经纠结于是否枚举对角线。但是细想一下，包含对角线的正方形其实应经在枚举过程中出现了，故不需要重复考虑。枚举时要考虑两个方向的正方形。最后每个正方形枚举了4次。关于哈希函数的选
2015-10-13 22:37:12
280
原创 poj Snowflakes snow snowflakes 哈希
题意：给出n个雪花的六条边的长度，求出是否有两个相同的雪花，即边的大小和顺序相同。因为只要考虑是否存在相同的雪花，结合雪花边长可能较大的条件，考虑哈希。需要顺时针和逆时针枚举比较。
2015-10-10 21:52:27
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原创 POJ 3026 Borg Maze 最小生成树
这道题大意很好懂，有一个S点和不超过100个A点，求这些点构成的完全图的最小生成树。题意虽简单，但实现略繁琐，而且Input有陷阱。在n，m这一行行末会有多余空格。。。（无聊。）然后用了prim来做（代码有点乱）。#include #include #include #include #include using namespace std;const int max
2015-10-03 23:59:48
276
原创 HDU 1258 Sum it up
题意：给出n个正整数，x[1..n]，对于给定的t（t < 1000），找出所有满足和为t的x的组合。降序输出。
2015-09-17 16:54:56
336
原创 hdu 2795 Billboard
题意：有一块黑板h * w，可以往上面贴广告，优先顺序是先尽量靠上，再尽量靠左。广告的参数为1 × w[i]，求每张广告贴的高度。分析：开始居然被h骗了。。一开始半天都没想好怎么做。 在h上建树，维护区间最大值，即区间能放的最大广告牌宽度。需要插入广告时，先query一次w[i]，判断是否有剩余宽度，如果有，返回的节点值k为尽量靠上（左）的，update（k, dat[k] - w[i]
2015-07-29 22:59:21
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原创 hdu 1754 I hate it
此题也是一眼就能看出来的线段树。。。但是爆栈了。记得hdu能加预编译命令扩栈，然后上网找到了。#pragma comment(linker, "/STACK:1024000000,1024000000")#include #include #include #include #include #include #include #include #include usi
2015-07-29 00:58:34
271
原创 poj 2253 Frogger和 poj 2485 Highways
这两题有着相似的地方。都只求所有路径中的最长边的最小值。不同之处在于Frogger求的是固定起点和终点，而Highways是所有路径。所以Highways可以用最小生成树来求。而Frogger中起点和终点可以当成一组特值，所以可以考虑先求任意两点间的最小最长边，可以由此想到Warshell-Floyd算法，但是需要修改。当路径经过k时，分别考虑[i, k] 和 [k, j] 的最小最长边，
2015-07-23 22:47:25
244
原创 poj 2385 Apple Catching
分析：题意大致为：两棵树每分钟仅有一棵会掉果子，设为Tree 1, Tree 2。可用数组记录，记为a[i][j] = 1或0。找时刻为 j 的状态，此时在树 i 上，可移动 k 次，即d[i][j][k]。 推测如何从前面的状态转移到此状态。前一状态有两种，即在j - 1时刻，这头牛要么在tree 1，要么在tree 2。可得状态转移方程：d[i][j][k] = max(d[i]
2015-06-19 00:24:05
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原创 UVa439 Knight Moves
Knight MovesTime Limit: 3000MS Memory Limit: Unknown 64bit IO Format: %lld & %lluDescriptionA friend of you is doing research on the Traveling Knight Problem (T
2015-04-08 20:52:40
342
原创 UVa Parentheses Balance
Parentheses BalanceTime Limit: 3000MS Memory Limit: Unknown 64bit IO Format: %lld & %llu
2015-04-08 20:43:08
320
原创 UVa 12504 Updating a Dictioinary
题目：UVa 12504 Updating a Dictionary分析：键值对可用map来处理，即map[ key ] = value。因为有三种情况要讨论，股分别用set保存对应变化的键值，顺便排序。读入需要注意，通过冒号来区分key和value，不同键值对用 ',' 和 '}‘区分，而 '}' 代表此行的读入停止。代码：#include #include #inc
2015-03-21 19:55:14
443
原创 UVa 1595 Symmetry
SymmetryTime Limit: 3000MS Memory Limit: Unknown 64bit IO Format: %lld & %lluSubmit StatusDescriptionThe figure shown on the left is left-right symmetric as
2015-03-15 17:44:52
340
原创 UVa 1593 Alignment of Code
Alignment of CodeTime Limit: 3000MS Memory Limit: Unknown 64bit IO Format: %lld & %lluSubmit StatusDescriptionYou are working in a team that writes Incredib
2015-03-15 16:24:56
320
原创 UVa 10763 Foreign Exchange
Problem EForeign ExchangeInput: standard inputOutput: standard outputTime Limit: 1 secondYour non-profit organization (iCORE - international Confederation of Revolver Enthusiasts) coordi
2015-03-15 15:25:36
276
原创 UVa 10935 Throwing cards away 1
Throwing cards away ITime Limit: 3000MS Memory Limit: Unknown 64bit IO Format: %lld & %lluSubmit StatusDescriptionProblem B: Throwing cards away IGiven is
2015-03-15 14:57:56
358
原创 UVa 1594 Ducci Sequence
Ducci SequenceTime Limit: 3000MS Memory Limit: Unknown 64bit IO Format: %lld & %lluSubmit StatusDescriptionA Ducci sequence is a sequence of n-tuples of int
2015-03-15 13:34:11
325
原创 UVa 232 Crossword Answers
232 - Crossword AnswersTime limit: 3.000 seconds A crossword puzzle consists of a rectangular grid of black and white squares and two lists of definitions (or descriptions).One list of def
2015-02-24 12:28:49
330
原创 zoj 1383 Binary Numbers
Binary NumbersTime Limit: 2 Seconds Memory Limit: 65536 KBGiven a positive integer n, print out the positions of all 1's in its binary representation. The position of the least significan
2015-02-11 14:16:33
347
原创 zoj 1331 Perfect Cubes
Perfect CubesTime Limit: 10 Seconds Memory Limit: 32768 KB For hundreds of years Fermat's Last Theorem, which stated simply that for n > 2 there exist no integers a, b, c > 1 such that
2015-02-05 15:22:35
334
原创 zoj 2423 Fractal
FractalTime Limit: 2 Seconds Memory Limit: 65536 KBA fractal is an object or quantity that displays self-similarity, in a somewhat technical sense, on all scales. The object need not exhi
2015-02-04 16:22:00
474
原创 zoj 1251 Box of Bricks
Box of BricksTime Limit: 2 Seconds Memory Limit: 65536 KBLittle Bob likes playing with his box of bricks. He puts the bricks one upon another and builds stacks of different height. ``Look
2015-02-04 16:13:02
297
原创 zoj 1242 Carbon Dating
Carbon DatingTime Limit: 2 Seconds Memory Limit: 65536 KBUntil the second half of the 20th century, determining the actual age of an archaeological find had been more or less a matter of
2015-02-04 15:35:34
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原创 zoj 1241 Geametry Made Simple
Geometry Made SimpleTime Limit: 2 Seconds Memory Limit: 65536 KBMathematics can be so easy when you have a computer. Consider the following example. You probably know that in a right-angl
2015-02-04 15:26:52
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原创 zoj 1240 IBM Minus One
IBM Minus OneTime Limit: 2 Seconds Memory Limit: 65536 KBYou may have heard of the book '2001 - A Space Odyssey' by Arthur C. Clarke, or the film of the same name by Stanley Kubrick. In i
2015-02-04 15:21:42
302
原创 zoj 1037 Gridland
GridlandTime Limit: 2 Seconds Memory Limit: 65536 KBBackgroundFor years, computer scientists have been trying to find efficient solutions to different computing problems. For some of
2015-01-30 15:05:32
294
原创 zoj 1067 Color Me Less
Color Me LessTime Limit: 2 Seconds Memory Limit: 65536 KBProblemA color reduction is a mapping from a set of discrete colors to a smaller one. The solution to this problem requires th
2015-01-30 14:55:29
429
原创 zoj 1115 Digital Roots
Digital RootsTime Limit: 2 Seconds Memory Limit: 65536 KBBackgroundThe digital root of a positive integer is found by summing the digits of the integer. If the resulting value is a si
2015-01-30 14:49:58
329
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