maxe

TV-MAXE is an online live TV software similar to Sopcast that supports GStreamer, VLC, and HTTP remote control. It uses Sopcast's P2P technology, supports mms, sop, and network streams, and supports TV station lists in multiple countries and regions. The latest available version is TV-Maxe 0.09, which supports stream recording, better support for m3u8 files, an improved GUI and other fixes. Ubuntu 14.04 do

Bsgs#include #include#include#include#include#definell Long Longusing namespacestd;Const intMaxint= ((1 -)-1)*2+1;inta,b,c;structhashmap{Static Const intHa=999917, maxe=46340; intE,lnk[ha],son[maxe+5],nxt[maxe+5],w[maxe+5]; intTop,stk[maxe+5]; voidClear () {e=0; while(top) l

' otherwise. Sample Input 13 31 22 33 1 Sample Output Yes SourcePOJ Monthly -- 2006.02.26, zgl twb A directed graph is given to determine whether u and v can reach v or u at any two points in the graph. If Yes, Yes is output and No is output. Note that here is or is not. If yes, you can directly determine whether the entire graph is a strongly connected component. This is much simpler, the idea of this question is to first scale down the entire graph (any two points in each strongly connected c

Main topicGives an no-map to find the maximum flow of a graph of all the shortest circuits from 1 to n on this graph.IdeasWant to let everybody stack template also not take so lazy.Remember to open long long on the line.CODE#define _crt_secure_no_warnings#include #include #include #include #include #define MAXP 10010#define Maxe 1000010#define S 0#define T (MAXP-1)#define INF 0X3F3F3F3F3F3F3F3FLLusing namespace STD;structmaxflow{intHEAD[MAXP], total;i

Main topicGiven a graph, each person can only do once a day plane. Now give the starting point, the end point, and the number of people to go, as well as the number of restrictions on each route, and ask at least how many days the slowest person reaches the finish line.Ideasis obviously the network flow model, as to how to verify, in fact, even two points are not used, enumerate the minimum number of days, and then each add a layer of edge to verify on the line.CODE#define _crt_secure_no_warning

#include#include#include#include#defineINF10000000000000007#definemaxn1010#definemaxm10010#definemaxe25000#definelllonglongusingnamespacestd;llc[maxm],v[maxn],w[maxe],cst[maxe],dis[maxn];int n,m,s,t,e,a[maxm],b[maxm];intfa[maxe],next[maxe],link[maxn],son[maxe],rec[

TvMaxe is a graphical interface of SopCast. It uses the P2P Technology of SopCast to support mms, RMB, and network streams. By default, TV stations in Spain, France, UK, and Denmark are listed. New version of TV-Maxe: HTTPremotecontrol: TVMaxealreadyhadsupportforremotecontrolviainfrared, butthelatestversio TV Maxe is a graphical interface of SopCast. It uses the P2P Technology of SopCast to support mms, RMB

Problem:Given A stringS, find the longest palindromic substring inS. Assume that maximum length ofSis, and there exists one unique longest palindromic substring.Solution:find palindrome strings in the middle of each character, then record the largest palindrome string, time complexity O (n^2)Main topic:returns the longest string of strings, given a string. Problem Solving Ideas:the easiest way to think of brute force, determine each substring, and then see if it is a palindrome string, time comp

Simple 2-sat.1 /*3678*/2#include 3#include 4#include 5#include 6#include 7 using namespacestd;8 9 #defineMAXN 1005Ten #defineMaxe 1000005 One Atypedefstruct { - intV, next; - } edge_t; the - intpre[maxn*2], low[maxn*2]; - intbn[maxn*2]; - intDfs_clock, block; + ints[maxn*2], top; - inthead[maxn*2], L; + edge_t E[maxe]; A intN, M, N2; at intA[maxe], B[maxe],

Let's start with a little recap, we're already going to find a linear equation (including its special case multiplication inverse)We will also perform a fast algorithm for power modulus (modulo is the Fermat theorem for prime numbers with Euler's theorem for general conditions)We also extend the expansion of Euler's theorem to solve the situation where the exponent is particularly large.For the modular linear equations, the modulus coprime is used directly with Sun Tzu's theoremWhen modulus is n

modified into pointersi =p; } Heap[i]=x; } voidpop () {Tx = heap[sz--]; inti =1; while((i1) SZ) { intA = i1, B = i1|1; if(Bb; if(! CMP (heap[a],x)) Break; Heap[i]=Heap[a]; I=A; } Heap[i]=x; } Toperator[](intx) {returnheap[x];};Const intMAXV = +, Maxe =4e4;intHd[maxv],to[maxe],nx[maxe],ec,cap[maxe],cost[

Poj_2396 The first time I came into contact with the Network Flow Problem of the upper and lower bounds, I felt that I had to follow the routine. The specificAlgorithmSee http://blog.csdn.net/water_glass/article/details/6823741. # Include # Include String . H> # Include # Define Maxn 210 # Define Maxm 30 # Define Maxv 230 # Define MaxE 24900 # Define INF 0x3f3f3f Int N, m, low [maxn] [maxm], high [maxn] [maxm], R [maxn], C [maxm]; Int S, T, S

Because the knight who hates each other cannot be adjacent, the knight can be connected to the non-side, the meeting request is odd, the problem is to seek no more than the number of nodes on a simple odd circle.Tarjan directly set previously written, the result is wrong, to pay special attention to the side to join the stack of time ... I don't understand the algorithm.#include using namespacestd;#defineBug (x) coutConst intMAXV = ++5;Const intMaxe = MAXV*MAXV;//Open Smallintdfn[maxv],low[maxv]

Few points, sort by edge weights, enumerate enumerations L and R, check connectivity. Once connected, update the answer.Judge even General can O (1), before I was O (n) sentenced, previously written over, and later wrote T ... #include using namespacestd;Const intMAXN =101;Const intMaxe = maxn*maxn>>1;intn,m;intU[maxe],v[maxe],w[maxe];intPa[maxn];inlineBOOLcmpint

There are n jobs, M machines, and you can rent or buy them for every kind of machine. Each work consists of several processes, each of which requires a machine to be completed, and you can do so by purchasing or renting a machine. Now give these parameters, for maximum profit input first row gives n,m (1#include using namespacestd;Const intINF =0x3f3f3f3f;#defineRep (I, J, K) for (int i = j; I Const intMAXN = -+ -;Const intMAXV =2400+ -, Maxe =3000000

namespace std;// Typedef long LL;// Typedef _ int64 LL;// Typedef long double DB;// Typedef unisigned _ int64 LL;// Typedef unsigned long ULL;# Define EPS 1e-8# Define maxn1600# Define MAXE 300000# Define INF 0x3f3f3f# Define PI acos (-1.0)// # Define MOD 99991.// # Define MOD 99990001.// # Define MOD 1000000007.# Define max (a, B) (a)> (B )? (A) (B ))# Define min (a, B) (a) # Define max3 (a, B, c) (max (a, B), c ))# Define min3 (a, B, c) (min (a, B)

Python implementation of the harris algorithm, harrispython Harris is the most common feature detection algorithm. First file harris. py Algorithm for testing the second file from PIL import Imagefrom numpy import *import harrisfrom pylab import *from scipy.ndimage import filtersim=array(Image.open('33.jpg').convert('L'))harrisim=harris.compute_harris_response(im)filtered_coords=harris.get_harris_points(harrisim)harris.plot_harris_points(im,filtered_coords) How to Implement the DES alg

Pinch point +SPFA Shortest path, because there is no ring, so direct SPFA.1 ConstMaxe=1000001;2 type3Node=Record4 F,t:longint;5 end;6 varN,m,s,i,j,ans,cnt,num,u,x,dgr:longint;7h,he,dfn,low,q,d,v,va,bl:array[0.. Maxe] of Longint;8b,bi:array[0.. Maxe] of node;9f:array[0.. maxe* -] of Longint;Tenp:array[0.. Maxe] of Boole

Returns the maximum and minimum values of an array. Calculate the maximum and minimum values of the array. You can traverse the array and record the maximum and minimum values respectively. This method requires 2N comparisons. If you want to reduce the number of comparisons, You can traverse the array, compare Adjacent Elements, and put the greater margin of the adjacent elements behind the smaller ones in front. Select the maximum value from the greater limit to the maximum value of the entire

Time: 2016-04-08-18:05:09 Title number: [2016-04-08][poj][3159][candies] The main topic: N Children, m information, each message A, B, C, said B children get sugar can not be more than a child C, ask N children than 1 children can more than the maximum number of Analysis: Direct is to ask for 1 to n the shortest path (if not the minimum, then the other side, there is always a situation can not meet the most short-circuit this side of the situation,), run directly once Dijkstra Pr