usaco 5.4 Betsy’s Tour(插頭DP一條迴路)

Betsy's TourDon PieleA square township has been divided up into N2 square plots (1 <= N <= 7). The Farm is located in the upper left plot and the Market is located in the lower left plot. Betsy takes her tour of the township going from Farm to

zoj 3256 Tour in the Castle(插頭DP一條迴路+矩陣連乘)

Tour in the CastleTime Limit: 5 Seconds      Memory Limit: 32768 KBAfter the final BOSS is defeated, the hero found that the whole castle is collapsing (very familiar scene, isn't it). Escape from the castle is easy, just need to cross a few rooms.

POJ3189 Steady Cow Assignment 最大流+枚舉

建圖,牛到源點弧的容量為1,牛到牛棚弧的容量為1,牛棚到匯點弧的容量為相應棚的容量。枚舉,每次枚舉一個區間,如果該level區間內的圖,所求出的最大流等於牛的數量,則輸出區間的差值。Steady Cow AssignmentTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 4127 Accepted: 1444DescriptionFarmer John's N (1 <= N <= 1000) cows each

POJ2135 Farm Tour 最小費用最大流

每一條路只能走一次,求一條起點到終點再到起點最短的路。那麼給每一條路賦值容量為1,費用是距離,從起點開始用最小費用流增廣2次後的費用就是最小費用。這題就是最小費用最大流。 Farm TourTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 7904 Accepted: 2837DescriptionWhen FJ's friends visit him on the farm, he likes to show them

hdu 3377(插頭DP一條迴路固定端點)

 PlanTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 207    Accepted Submission(s): 54Problem DescriptionOne day, Resty comes to an incredible world to seek Eve -- The origin of life. Lilith,

hdu 1239 Calling Extraterrestrial Intelligence Again (暴力枚舉)

Calling Extraterrestrial Intelligence AgainTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 3993    Accepted Submission(s): 2097Problem DescriptionA message from humans to extraterrestrial

ZOJ 2900 Icecream(線段樹)

題意:給你一個長度為n的串,從中選出字串,使得長度至少為k,且相鄰的數值差的絕對值小於等於p,求這樣的字串的個數mod m的值地址:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=2900分析:我們可以馬上想到,以前i個數,長度為j的字串個數為f[ i ][ j ],那麼有f[ [i ][ j ]=sum{ f[ i' ][ j-1 ] }

HDU 1812 Count the Tetris (polya定理+高精度)

題目地址:http://acm.hdu.edu.cn/showproblem.php?pid=1812標準的polya定理問題。旋轉只有 0,90,180,270度三種旋法。旋0度,則置換的輪換數為n*n旋90度,n為偶數時,則置換的輪換數為n*n/4,n為奇數,則置換的輪換數為(n*n-1)/4+1旋180度,n為偶數時,則置換的輪換數為n*n/2,n為奇數,則置換的輪換數為(n*n-1)/2+1旋270度,n為偶數時,則置換的輪換數為n*n/4,n為奇數,則置換的輪換數為(n*n-1)/4+

POJ 3233 Matrix Power Series (矩陣+二分+二分)

題目地址:http://poj.org/problem?id=3233題意:給你一個矩陣A,讓你求A+A^2+……+A^k模p的矩陣值題解:我們知道求A^n我們可以用二分-矩陣快速冪來求,而當k是奇數A+A^2+……+A^k=A^(k/2+1)+(A+A^2+……A^(k/2))*(1+A^(k/2+1))當k是偶數A+A^2+……+A^k=(A+A^2+……A^(k/2))*(1+A^(k/2))可以在一次用二分。AC代碼:#include <iostream>#include &

hdu 4277 USACO ORZ (暴力+set容器判重)

USACO ORZTime Limit: 5000/1500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2291    Accepted Submission(s): 822Problem DescriptionLike everyone, cows enjoy variety. Their current fancy is new shapes for pastures.

poj 2420 A Star not a Tree?(類比退火求費馬點)

A Star not a Tree?Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 2519 Accepted: 1332DescriptionLuke wants to upgrade his home computer network from 10mbs to 100mbs. His existing network uses 10base2 (coaxial) cables that allow you to

最小樹型圖的求解與實現

轉自:http://www.zlinkin.com/?p=63  圖論是ACM競賽中比較重要的組成部分,其模型廣泛存在於現實生活之中。因其表述形象生動,思維方式抽象而不離具體,因此深受各類喜歡使勁YY的Acmer的喜愛。這篇文章引述圖論中有關有向圖最小產生樹的部分,具體介紹朱劉演算法的求解思路,並結合一系列coding技巧,實現最小樹型圖O(VE)的演算法,並在最後提供了該演算法的模版,以供參考。  關於最小產生樹的概念,想必已然家喻戶曉。給定一個連通圖,要求得到一個包含所有頂點的樹(原圖的子圖

hdu 4435 charge-station  (貪心+圖論)

charge-stationTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 587 Accepted Submission(s): 291Problem DescriptionThere are n cities in M^3's empire. M^3 owns a palace and a car and the palace

POJ3264 Balanced Lineup

這道題很基礎,幾乎就是一個ST演算法的模板。第一次接觸ST演算法可以用這道題來練習演算法。 Balanced LineupTime Limit: 5000MS Memory Limit: 65536KTotal Submissions: 23289 Accepted: 10827Case Time Limit: 2000MSDescriptionFor the daily milking, Farmer John's N cows (1 ≤ N ≤ 50,000) always line up

POJ1465 Multiple BFS+同餘

給你一個數n 然後給你m個數,讓你求一個最小的數,這個數是n的倍數,並且由題目中提供的m個數組成。用BFS。此題可以用同餘判斷的方法來剪枝。假如 A%X  ==  B%X  (設A<B)那麼 (A*10+Ki)%X==(B*10+Ki)%X所以A,B之中我們只要取前面的A就行,因為題目要取最小的數,通過同餘我們可以知道,A和B這兩個數在末尾添加任何相同的數MOD

hdu 2993 MAX Average Problem(DP+斜率最佳化入門題)

MAX Average ProblemTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3800    Accepted Submission(s): 980Problem DescriptionConsider a simple sequence which only contains positive integers as a1,

我的跳舞鏈 Dancing Links 模板

第一個模板——精確覆蓋問題題目:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=3038為什麼選擇這題呢,因為它既可以當作數獨模板又能當成DLX模板,不是一舉兩得嗎^ ^#include<cstdio>using namespace std;const int N=16;const int mm=N*N*N*4+N*N*4+N;const int mn=N*N*N+N;int

ZOJ2112 Dynamic Rankings 線段樹套平衡樹

簡單樹套樹區間單點修改區間K值查詢線段樹套TREAPDynamic RankingsTime Limit: 10 Seconds      Memory Limit: 32768 KBThe Company Dynamic Rankings has developed a new kind of computer that is no longer satisfied with the query like to simply find the k-th smallest number of

網路流之–混合圖的歐拉迴路

基礎知識    歐拉迴路是圖G中的一個迴路,經過每條邊有且僅一次,稱該迴路為歐拉迴路。具有歐拉迴路的圖稱為歐拉圖,簡稱E圖。    無向圖中存在歐拉迴路的條件:每個點的度數均為偶數。    有向圖中存在歐拉迴路的條件:每個點的入度=出度。    歐拉路徑比歐拉迴路要求少一點:    無向圖中存在歐拉路徑的條件:每個點的度數均為偶數或者有且僅有2個度數為奇數的點。    有向圖中存在歐拉路徑的條件:除了2個點外,其餘的點入度=出度,且在這2個點中,一個點的入度比出度大1,另一個出度比入度大1。  

ZOJ3645 BiliBili 高斯消元法

之前數值計算學過,可是好久不複習給忘記了。於是拿出出來複習。終於搞定。題目意思很明顯給你一個方程組 讓你求(x1,x2,x3,x4,x5,x6,x7,x8,x9,x10,x11)但是方程組的形式是一個二次方程組(ai1-x1)^2 + (ai2-x2)^2 +(ai3-x1)^2 + (ai4-x2)^2 +(ai5-x1)^2 + (ai6-x2)^2 +(ai7-x1)^2 + (ai8-x2)^2 + (ai9-x2)^2 +(ai10-x1)^2 + (ai11-x2)^2  =

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