其他

其實最開始看沒啥思路，還以為是DP什麼的。冷靜分析一下，我們可以把橫豎行分開來看，那麼有一個傘兵就在左右臉上一條線（流量無窮），這不是個二分圖嗎？！然後就很自然想到加上一個源點一個匯點就成了網路流求最小割，也就是最大流的問題上來。NICE。#include "stdio.h"#include "iostream"#include "stdlib.h"#include "queue"#include "math.h"#include "string.h"using namespace

這題第一感覺就是Floyd：方便快捷，也很直接。sum+=d[1][i]+d[i][1]  (i---- 1->n);但顯然悲劇逾時，然後就想到Floyd的最佳化演算法Spfa。一般的都會用上隊列，比如vector確實好用，但這題太坑，卡vector逾時，然後聽取了其他人的想法用鄰接表解決了。#include<stdio.h>#define MAXN 1000010#define MOD 1000007#define inf 10000000000000LLtypedef

Dinic 演算法的基本步驟為:1) 計算殘餘網路的層次圖。我們定義 h[i] 為頂點 i 距離源 S 所經過到最小邊數，求出所有頂點的 h 值，h[] 值相同的頂點屬於同一層，這就是網路的層次圖。2) 在層次圖上進行 BFS 增廣，直到不存在增廣路徑。這時求得的增廣路徑上頂點是分層的，路徑上不可能存在兩個頂點屬於同一層，即 h[i]== h[j] (i!= j )。同時，求得層次圖後，我們可以在層次圖上進行多次增廣。3) 重複 1 和

在SAP演算法中，我們定義每個頂點的距離標號(Distance Labels)，即殘留網路中這個點到匯點的距離。然後，我們只在距離標號相鄰的點間尋找增廣路徑。如果從一個點出發沒有容許邊，就對重標記並回溯（類似預流推進）。首先介紹一些定義：殘留容量：容量－流量。即，殘留容量r[i,j]=c[i,j]-f[i,j]。如果只求流值，則在網路流演算法中只需記錄殘量，而不必記錄容量和流量。殘留邊、殘留網路：有殘留容量的邊成為殘留邊，由網路的點集、殘留邊集構成殘留網路。距離標號

這題又做了大半天。。題目如下DescriptionOnce again, James Bond is on his way to saving the world. Bond's latest mission requires him to travel between several pairs of cities in a certain country. The country has N cities (numbered by 1, 2, . . ., N), connected by 

好長時間沒寫部落格了，一想起“水可三日不飲，飯可七日不食，而程不可一日不編”，還是回顧一下最近的內容，下午打了場多校說下這道A下的第一題。Problem DescriptionThe I-number of x is defined to be an integer y, which satisfied the the conditions below:1. y>x;2. the sum of each digit of y(under base 10) is the

題目是這樣的Problem DescriptionClaire and her little friend, ykwd, are travelling in Shevchenko's Park! The park is beautiful - but large, indeed. N feature spots in the park are connected by exactly (N-1) undirected paths, and Claire is too tired to

Tree RecoveryTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 4585 Accepted: 3061DescriptionLittle Valentine liked playing with binary trees very much. Her favorite game was constructing randomly looking binary trees with capital letters in

題目大意：給定一個由SNWE幾個字母組成的地圖,這四個字母代表著四個方向，給定起點，讓你判斷是否能走出這個地圖，或走到某步的時候形成一個環。。。Sample Input3 6 5NEESWEWWWESSSNWWWW4 5 1SESWEEESNWNWEENEWSEN0 0 0Sample Output10 step(s) to exit3 step(s) before a loop of 8

DescriptionGiven a positive integer n, write a program to find out a nonzero multiple m of n whose decimal representation contains only the digits 0 and 1. You may assume that n is not greater than 200 and there is a corresponding m containing no

StuPIdTime Limit: 3000MS Memory Limit: 65536KTotal Submissions: 5111 Accepted: 2631DescriptionBackground At DUT, the Dreamland University of Technology, all students have personal id, numbers with six or seven digits. But they're not just any

An Easy ProblemTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 5783 Accepted: 3344DescriptionAs we known, data stored in the computers is in binary form. The problem we discuss now is about the positive integers and its binary form. Given

Power StringsTime Limit: 3000MS Memory Limit: 65536KTotal Submissions: 13765 Accepted: 5738DescriptionGiven two strings a and b we define a*b to be their concatenation. For example, if a = "abc" and b = "def" then a*b = "abcdef". If we think of

棋盤問題Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 9462 Accepted: 4600Description在一個給定形狀的棋盤（形狀可能是不規則的）上面擺放棋子，棋子沒有區別。要求擺放時任意的兩個棋子不能放在棋盤中的同一行或者同一列，請編程求解對於給定形狀和大小的棋盤，擺放k個棋子的所有可行的擺放方案C。Input輸入含有多組測試資料。 每組資料的第一行是兩個正整數，n

Autumn is a GeniusTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 6040 Accepted: 1269DescriptionJiajia and Wind have a very cute daughter called Autumn. She is so clever that she can do integer additions when she was just 2 years old!

/*邊聯通度的題；給定n個門，每個門串連兩個房間，門面向其中的一個房間x，從這個房間x任意情況都可以到另外一個房間y,y只能在門開著的時候可以進入x,其中有些房間有入侵者，現在求解至少關幾個門，可以保護某個房間的安全對於x,y之間面向x的門，連x到y的邊，流量無窮，y到x的邊，流量1串連源點到有入侵者的點，流量無窮，要保護的點作為匯點如果最大流為無窮，則不存在保護方案，反之，最大流即是最少要關的門的數量*/#include <stdio.h>#include

Calling Extraterrestrial Intelligence AgainTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 8938 Accepted: 3585DescriptionA message from humans to extraterrestrial intelligence was sent through the Arecibo radio telescope in Puerto Rico on

馬和象的故事TimeLimit: 1 Second   MemoryLimit: 32 Megabyte Totalsubmit: 215   Accepted: 46   Special Judge

ebola.exe 遍曆硬碟和隨身碟檔案:如果尾碼名為exe，則刪除檔案。如果為c，則在其目錄下產生包含啟動ebloa.exe代碼的stdio.c標頭檔，並設其屬性為隱藏,修改檔案頭<stdio.h>為"stdio.c",並在c檔案中插入惡意函數。cpp和html的還沒找到惡意代碼，暫不做處理。如果尾碼名為gho，則刪除檔案。 修改註冊表:屏蔽安全模式,添加自啟動，修改鎖定首頁,禁止查看隱藏檔案,屏蔽exe檔案尾碼名,添加開機自啟動時可能會有殺軟的攔截導致失敗，應該再判斷下，

Agri-NetTime Limit: 1000MS Memory Limit: 10000KTotal Submissions: 15626 Accepted: 6327DescriptionFarmer John has been elected mayor of his town! One of his campaign promises was to bring internet connectivity to all farms in the area. He needs your