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[Abstract]: This paper introduces the DVB-based Data Broadcast System (IP over DVB) developed by the Blue topology company for cable TV networks, which focuses on the system structure, the front-end system structure and implementation, and the design and implementation of the backend data receiving card. Key parts of the system: Front-End Server software, multi-protocol wrappers, and data receiving cards ar

The access network system is widely used and has been well developed in various fields. It also plays an important role in cable TV return. After more than 20 years of development, China's cable TV network line is now more than 3 million kilometers in length, the optical fiber trunk line reaches 0.26 million kilometers

CATV Network is an efficient and inexpensive comprehensive network, it has a wide bandwidth, large capacity, multi-function, low cost, anti-interference ability, support a variety of services to connect the advantages of millions of households, its development for the development of Information superhighway laid a foundation. In recent years in China CATV Network development, the country has built more than 1500 cable

Cable TV NetworkAbstract: Give a graph with n points vertex, give M edge. Find the point of connectivity KAlgorithm: Each vertex v is split into V ' V ', and V '-->v ' has a capacity of 1. For edges in the original (u,v) edge u '--->v ' V '-->u '. P (u,v) for each pair of points, with u as source and V as meeting point.We just need to fix one vertex and enumerate the other sinks.1#include 2#include 3#inclu

(inti = head[x]; I i =Edge[i].next) { intv =edge[i].to; if(EDGE[I].W dep[v] = = Dep[x] +1) {k=Dinic (V,min (rest, EDGE[I].W)); if(!k) Dep[v] =0; EDGE[I].W-=K; Edge[i^1].W + =K; Rest-=K; } } returnFlow-rest;}BOOLBFs () {memset (DEP,0,sizeof(DEP)); Queueint>Q; Q.push (s); Dep[s]=1; while(!Q.empty ()) { intx =Q.front (); Q.pop (); for(inti = head[x]; I i =Edge[i].next) { intv =edge[i].to; if(EDGE[I].W !)Dep[v]) {Q.push (v); DEP[V]= Dep[x] +1; if(v =

Links:http://poj.org/problem?id=1966Test Instructions: an undirected graph, n points, M edges, to find the vertex connectivity of this graph.idea: vertex connectivity, that is, the number of cut points in the minimum cut point set, the general method of finding the vertex connectivity of undirected graphs is to transform to the minimum cut of the network flow.Build diagram:(1) The original image of each point I split, split for I ' and I ', I ' to I ', an arc capacity of 1.(2) for edges present

Cable TV NetworkTime limit:3000ms 64bit IO Format:%lld %llu The point connectivity of undirected graphs is obtained, that is to say, to find the minimum cut of any two points.But considering s set and T set, the middle is connected by several edges, these edges are the smallest cuts of s to T, assuming that this is the answer, then finding that there is no need to enumerate all the points, but to fix a poin

Comparison of IP transmission technology in cable TV networks Abstract: IP transmission technology in cable TV networks includes three forms: IP over ATM, IP over SDH, and IP over WDM, this article introduces and compares the three IP Address Transmission Technologies in detail. Key words: IP technology,

win8.1 system notebook connection use HDMI HD cable to connect to TV no sound how to solve? win8.1 system notebook has an HDMI interface, as long as the use of HDMI high-definition cable can be connected to the monitor, LCD TV screen display, so you can operate under the large screens. But some win8.1 system friends sa

POJ 1966 Cable TV NetworkLinks: http://poj.org/problem?id=1966Test Instructions: in a cable TV network, the repeater connection is bidirectional. If there is at least one route between any two repeaters in the network, the repeater network is called connected, or the repeater network is not connected. An empty network,

Cable TV Network Time Limit: 1000MS Memory Limit: 30000K Total Submissions: 4234 Accepted: 1989 DescriptionThe interconnection of the relays in a cable TV network is bi-directional. The network is connected if there are at least one interconnection path be

Cable TV Network Time Limit: 1000MS Memory Limit: 30000K Total Submissions: 4690 Accepted: 2170 DescriptionThe interconnection of the relays in a cable TV network is bi-directional. The network is connected if there are at least one interconnection path be

2016-05-31 13:25:45Topic Link: Rokua P1273 cable TV networkMain topic:Take as many leaf nodes as possible on a given tree of weights, making sigma (val[selected leaf node])-sigma (cost[)) >=0Solution:Tree-like DP backpack DPDP[I][J] represents the maximum profit obtained by selecting J leaf nodes for the subtree of the root of the I node.Transfer equationDp[i][j]=max (Dp[i][j],dp[i][j-k]+dp[son][k]-cost[son

while(Tell ()) { the for(intI=1; i) -cr[i]=H[i]; Wu while(t=Zeng (S,inf)) -r+=T; About } $ returnR; - } - intMain () { - while(SCANF ("%d%d", n,m)! =EOF) { Amemset (H,0,sizeof(h)); +memset (NXT,0,sizeof(NXT)); thetot=1; -ans=INF; $ for(intI=1; i) theAdd (I,i+n,1); the for(intI=1; i){ thescanf"(%d,%d)",aa,BB); theAdd (aa+n+1, bb+1, INF); -Add (bb+n+1, aa+1, INF); in } the for(intI=1; i) the for(intj=1; j){ About

+ =tmp; the if(Cost = =limit) the Break; +}Else { -DIS[V] =-1; the }Bayi } the } the returnCost ; - } - intDinic () { the intAns =0; the while(BFS ()) { theAns + =DFS (s, INF); the } - returnans; the } the intMain () the {94 intN, m, x, y; the while(~SCANF ("%d%d", n, m)) { the for(inti =0; i) { thescanf"(%d,%d)", edge[i][0], edge[i][1]);98 } Abouts =N; - intAns =inf;101

The main topic: to a N-point, M-edge of the graph, the minimum point cut set of the cardinality.Problem Analysis: The base of the minimum point cut set of the graph can be changed to the minimum cut. Considering the graph of single source s single-sink T, if a minimum point set is required, so that the graph is no longer connected after the point set is removed (the number of connected components is increased), simply split each point into two (in and out points), and with a capacity of 1 arc, t