hdu Non-negative Partial Sums(單調隊列)

Non-negative Partial SumsTime Limit : 6000/3000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)Total Submission(s) : 6   Accepted Submission(s) : 1Font: Times New Roman | Verdana | GeorgiaFont Size: ← →Problem DescriptionYou are given a

Poj A Simple Problem with Integers(lazy線段樹)

A Simple Problem with IntegersTime Limit : 10000/5000ms (Java/Other)   Memory Limit : 262144/131072K (Java/Other)Total Submission(s) : 40   Accepted Submission(s) : 13Problem DescriptionYou have N integers, A1, A2, ... , AN. You need to deal with

寒假前刷題(1)

這是一道關於二叉樹的題,具體的題目是杭電的1710 ,就是通過二叉樹的前序和後序來推出二叉樹的中序序列,一道很簡單的題,不過筆者還是太笨了,花了好久才搞定。具體的思想就是先通過一個前序序列找到它的根節點,在通過找到根節點在中序序列裡的位置來得出,該位置前的全為左子樹的結點,後面的全為右子樹的結點,在通過層層遞迴得出。代碼如下:#include<stdio.h>#include<string.h>#include<iostream>using

求最近點對的基礎演算法

    近期演算法課上,剛剛學習了 關於最近點對的相關知識,目前只是參看了大牛的思想 ,寫了個最基本的裸最近點對。下面是兩種方式    蠻力法:#include<cstdio>#include<cstdlib>#include<cstring>#include<cmath>using namespace std;struct p{ int x; int y;};double ClosestPoint1(int n,p a[],int

hdu 3172 && hdu 3047

hdu 3172並查集,大水題直接代碼:#include<iostream>using namespace std;#include<cstdio>#include<map>const int maxn=100005;int pre[maxn],val[maxn];map<string,int> m;inline int input(){ char c; int ret=0; c=getchar(); while(c<

CF 179(div2)D(floyed)

D. Greg and Graphtime limit per test3 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputGreg has a weighed directed graph, consisting of n vertices. In this graph any pair of distinct vertices has an edge between them

CF 172(div2) D(單調隊列)

D. Maximum Xor Secondarytime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputBike loves looking for the second maximum element in the sequence. The second maximum element in the sequence of distinct

關於floyd的一點 總結

floyd算是求最短路徑裡面的最簡單的一種演算法了。  它一般用在求一個有向網(或無向網),對每一個頂點vi不等於vj,要求求出vi與vj之間的最短路徑和最短路徑的長度。如果用dijkstra的話要迴圈n次,用floyd雖然時間複雜性並未減少,同樣是 o(n^3),但是寫起來更加簡單。這個裡面沒什麼特別好講的地方,值得注意的是他的兩個遞推公式。A^k[i][j]=min{A^(k-1)[i][j],A^(k-1)[i][k]+A^(k-1)[k][j]}

CF 189 div2 D

D. Psychos in a Linetime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputThere are n psychos standing in a line. Each psycho is assigned a unique integer from1 to n. At each step every psycho who has

CF 154 div2 B(dp)

B. Physics Practicaltime limit per test1 secondmemory limit per test256 megabytesinputinput.txtoutputoutput.txtOne day Vasya was on a physics practical, performing the task on measuring the capacitance. He followed the teacher's advice and did as

關於dijkstra的一點總結

最近的一段時間都在刷關於dijkstra的東西 ,在參看了白書等參考之後 ,我個人做了點小總結。首先要明確的是dijkstra的最主要用途是求單源最短路徑的(權值非負)。所謂單源最短路徑就是固定一個頂點為源點,然後求源點到其他點的最短路徑。(需要特別注意的是

uva 133 救濟金髮放

#include<cstdio>#include<cstring>#include<cstdlib>#include<iostream>using namespace std;int pre[25],next[25];int N,m,n;int pos1,pos2;//移除結點void remove(int n){ next[pre[n]]=next[n];//第二個人的下一個位置

Hdu 2955 Robberies//01背包

題目描述:Roy去搶銀行,如何在不被抓住的情況下搶到最多的錢。給出搶每個銀行能搶到的錢和被抓到的機率。分析:剛開始以為機率只有兩位小數,乘以100直接做的,結果探索資料不是這樣的,果斷wa了。然後一直在這裡糾結,背包容量不為整數怎麼辦。無奈之下看一下別人的解題報告,發現可以讓搶的錢看做背包容量,被抓住的機率為背包價值。(趕腳自己在死學習,o(╯□╰)o)。但是自己又想錯了,直接將每次搶銀行沒被抓住的機率想加,懂點機率的人都知道這是很愚蠢的想法(o(╯□╰)o)。下面是代碼:/*Hdu 2955

Poj 2082(單調棧)

Terrible SetsTime Limit: 1000MS Memory Limit: 30000KTotal Submissions: 2778 Accepted: 1413DescriptionLet N be the set of all natural numbers {0 , 1 , 2 , . . . }, and R be the set of all real numbers. wi, hi for i = 1 . . . n are some elements in N,

最簡單的暴力求解演算法 簡單枚舉

題目大意是:輸入正整數n,按從小到大的順序輸出所有形如abcde/fghij=n的運算式,其中a-j恰好為數字0-9的一個排列,2=<n<=79.楊麗輸入:62輸出:79546/01283=6294736/01528=62 //這是一道非常沒含量的簡單的枚舉的題,所以我沒把他算在寒假前刷題裡,之所以把他寫出來是因為這是我第一次真正的開始系統的看關於暴力演算法的東西,暑假的時候打了醬油,現在就是在贖罪,老是聽十元大師說暴力辦法解決這道題,感覺暴力似乎是個很牛逼的東西。今天就寫個最傻逼的

poj 3009 搜尋

Curling 2.0Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 8582 Accepted: 3603DescriptionOn Planet MM-21, after their Olympic games this year, curling is getting popular. But the rules are somewhat different from ours. The game is played

poj 1611 並查集

The SuspectsTime Limit: 1000MS Memory Limit: 20000KTotal Submissions: 18435 Accepted: 8920DescriptionSevere acute respiratory syndrome (SARS), an atypical pneumonia of unknown aetiology, was recognized as a global threat in mid-March 2003. To

poj 2263&& zoj1952 floyd

Fiber NetworkTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 2725 Accepted: 1252DescriptionSeveral startup companies have decided to build a better Internet, called the "FiberNet". They have already installed many nodes that act as routers

poj 2253&zoj 1942

FroggerTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 20716 Accepted: 6741DescriptionFreddy Frog is sitting on a stone in the middle of a lake. Suddenly he notices Fiona Frog who is sitting on another stone. He plans to visit her, but

uva127 紙牌遊戲 (鏈表題)

                               最近非常的頹廢,好長時間都沒有刷題了,看了前面寫的部落格,感覺那時候寫下的目標全部廢了,不知道現在幡然醒悟,算不算晚,總而言之,希望能堅持下去吧,也希望這是一個好的開端,進擊的acmer!!!!!                           

總頁數: 61357 1 .... 7793 7794 7795 7796 7797 .... 61357 Go to: 前往

聯繫我們

該頁面正文內容均來源於網絡整理,並不代表阿里雲官方的觀點,該頁面所提到的產品和服務也與阿里云無關,如果該頁面內容對您造成了困擾,歡迎寫郵件給我們,收到郵件我們將在5個工作日內處理。

如果您發現本社區中有涉嫌抄襲的內容,歡迎發送郵件至: info-contact@alibabacloud.com 進行舉報並提供相關證據,工作人員會在 5 個工作天內聯絡您,一經查實,本站將立刻刪除涉嫌侵權內容。

A Free Trial That Lets You Build Big!

Start building with 50+ products and up to 12 months usage for Elastic Compute Service

  • Sales Support

    1 on 1 presale consultation

  • After-Sales Support

    24/7 Technical Support 6 Free Tickets per Quarter Faster Response

  • Alibaba Cloud offers highly flexible support services tailored to meet your exact needs.