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转载单调队列

一、什么是单调（双端）队列单调队列，顾名思义，就是一个元素单调的队列，那么就能保证队首的元素是最小（最大）的，从而满足动态规划的最优性问题的需求。单调队列，又名双端队列。双端队列，就是说它不同于一般的队列只能在队首删除、队尾插入，它能够在队首、队尾同时进行删除。【单调队列的性质】一般，在动态规划的过程中，单调队列中每个元素一般存储的是两个值：1、在原数列中的位置（下标）2

2012-08-08 10:45:26 515

原创 bnu 12872 Array Diversity

Enough with this Harry Potter, please! What are we, twelve-year olds? Let's get our teeth into some real pumpkin pasties -- oops, programming problems!Here we go!Let's define the diversity of a

2012-08-05 14:24:32 696

原创 hdu 4334 Trouble

Problem DescriptionHassan is in trouble. His mathematics teacher has given him a very difficult problem called 5-sum. Please help him.The 5-sum problem is defined as follows: Given 5 sets S_1,..

2012-08-04 00:01:15 504

原创 Codeforces Round #131 (Div. 2)

B. Hometasktime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputFurik loves math lessons very much, so he doesn

2012-08-01 16:07:12 931

原创 hdu 4320 Arcane Numbers 1

Problem DescriptionVance and Shackler like playing games. One day, they are playing a game called "arcane numbers". The game is pretty simple, Vance writes down a finite decimal under base A, and

2012-07-31 21:05:45 500

原创 hdu 4313 Matrix

Problem DescriptionMachines have once again attacked the kingdom of Xions. The kingdom of Xions has N cities and N-1 bidirectional roads. The road network is such that there is aunique path betw

2012-07-27 11:10:54 2234

原创 HDU 4311 Meeting point-1

Problem DescriptionIt has been ten years since TJU-ACM established. And in this year all the retired TJU-ACMers want to get together to celebrate the tenth anniversary. Because the retired TJU-ACM

2012-07-26 21:01:16 557

原创 UVALive 5865 Finding Bottleneck Shorstet Paths

题目链接：http://livearchive.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=3876不知道比赛的时候是怎么想的，居然没有想到并查集，肯定是太久没做，失去对并查集的敏感度了。。。。数据范围比较小，所有的边也只有100W而已，明显可以过的。感觉就是最小

2012-07-25 16:49:13 537

转载 KMP算法详解

转载自http://www.matrix67.com/blog/archives/115如果机房马上要关门了，或者你急着要和MM约会，请直接跳到第六个自然段。我们这里说的KMP不是拿来放电影的（虽然我很喜欢这个软件），而是一种算法。KMP算法是拿来处理字符串匹配的。换句话说，给你两个字符串，你需要回答，B串是否是A串的子串（A串是否包含B串）。比如，字符串A="I'm ma

2012-07-22 10:55:32 938

原创 HDU 4302 holedox eating

Holedox EatingTime Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 857 Accepted Submission(s): 292Problem DescriptionHoledox is a

2012-07-21 09:58:59 451

原创手写最大堆

#include#includeusing namespace std;int heap[1000010];int tot;#define lson ((rt)<<1)//注意加外括号#define rson (((rt)<<1)|1)void query(int rt){ heap[1]=heap[--tot]; while(1) { if(lson<=tot ||

2012-07-18 13:49:10 596

转载 CodeForces #113 Div.2 E

题目：http://www.codeforces.com/problemset/problem/166/E题意：给你一个四面体，上面有四个顶点，有一只蚂蚁站在点D上。现在给你一个数字n（n分析：n很大，有10的7次方，所以DFS是必定超时的，所以我们要寻求更好的方法来解决问题。f[i] 表示走i步回到起点的方法数。我们再定义一个变

2012-07-18 10:10:46 823

原创 Codeforces Round #113 (Div. 2) D. Shoe Store

D. Shoe Storetime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputThe warehouse in your shop has n shoe pairs.

2012-07-17 23:50:02 1193

原创 Codeforces Beta Round #86 (Div. 2 Only) B PFAST Inc.

B. PFAST Inc.time limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputWhen little Petya grew up and entered the uni

2012-07-14 23:29:07 1740

原创 Codeforces Beta Round #27 (Codeforces format, Div. 2) D Ring Road 2（二分图染色版）

D. Ring Road 2time limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputIt is well known that Berland has n cities, which form

2012-07-14 20:41:51 2116

原创 BNU 12889 Alice and Bob

Alice and BobTime Limit: 30000msCase Time Limit: 1000msMemory Limit: 65536KB 我不知道Alice和Bob到底发展到了哪一阶段，就像我不知道李雷最后有没有负了韩梅梅，同样的我也不知道奶牛Bessie是否逃出了Farmer John魔爪。撇开这些八卦不说，取石子游戏绝对是一款风靡ACM界的经

2012-07-14 16:00:06 820

原创 SPOJ Problem Set (classical) 11582. A Famous Grid

//螺旋数，很多时候不是你不会，是对未知的东西充满了恐惧//其实静下心来，这个螺旋数是完全可以写出来的，剩下的就是简单的bfs了//PS：图是无限大的，虽然x,y给出的范围是10000，也就是100*100的图，但是可以绕着走，所以要稍微开大一点儿#include#include#include#include#include#includeusing namespace st

2012-07-12 22:18:00 648

转载关于dijkstra

//v表示起始点，n表示总点数，pre表示上一个点，dis[i]表示当前起始点到i的最短路径，c为所保存的邻接矩阵void dijkstra(int n,int v,int pre[],int dis[],int c[][]){ bool vis[maxn];//是否用过 for(int i=1;i<=n;i++)//初始化，从起始点开始有路径相连的更新，所有点都未走过，距离更新 {

2012-07-12 20:46:18 380

原创 hdu 3118 【关于二分图】

/*根据题意，本题要求为图中不能出现奇数环由此想到二分图的性质：图中的点分为两个集合，一个集合中的点只有另一个集合有连边，而不与自身集合内的点有连边，因为只有两个集合，所以要形成环必须为偶数条边一共最多有15个点，所以最多有2^15次方种情况*/#include#include#include#includeusing namespace std;int mapp[17][1

2012-07-12 20:42:29 548

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