network graph visualization

D. There is a one-way network between schools, and each school receives a set of software that can be transmitted through a one-way network to the surrounding schools,Question 1: Initially, at least how many schools need to be distributed software, so that all schools in the network will eventually get the software.Question 2: At least a few transmission lines (e

them.InputThe input file consists of several blocks of lines. Each block describes one network. The first line of each block there is the number of places N N lines contains the number of a place followed by the numbers of some places to which t Here are a direct line from the this place. These at the very N lines completely describe the network, i.e., each direct connection of both places in the

the absence of a setExample: bzoj1497,bzoj2127,bzoj19342. Two points belong to the same set pay Val priceAt this time, the nature of the graph is a binary graph. (Face-beatingThen we can flip the nature of the two-part point, that is, the left point is selected, and the right point even T is selected.So after the transformation the problem became the first one. And just (I,j,val) (J,i,val)Example: bzoj1976

Topic Links:POJ1637Test instructionsA picture that gives a forward and a no-edge, asking if there is a Euler circuit that passes through all sides only onceProblem Solving Ideas:The solution of Euler circuit of mixed graph requires the use of network flow, and the concrete modeling method is as follows:1, first to all the non-directional direction, and then the statistics of all points in the degree and the

. #include Description there are n children in kindergarten who plan to vote to decide whether to take a nap. For them, this issue is not very important, so they decided to carry forward the spirit of humility. Although everyone has their own opinions, in order to take care of the ideas of their friends, they can also vote for the opposite of their own intentions. We define the number of conflicts in a vote as the total number of conflicts between good friends plus the number of people who

(intI=1; i0); for(intI=1; iintU=in (), V=in (); Add (U,v+n,inf), add (V+n,u,0); }}voidISAP () {memset(Dis,0,sizeof(dis));memset(Gap,0,sizeof(GAP)); for(intI=0; iintu=s,maxn=0, K=inf,v; gap[0]=t+1; Pre[s]=s; while(dis[s]BOOLf=0; while(!f) {f=1; for(intI=cur[u]; i!=-1; I=edge[i].next) {v=edge[i].v;if(edge[i].cap>0 dis[u]==dis[v]+1) {k=min (K,EDGE[I].CAP); Pre[v]=u; Cur[u]=i; U=v;if(u==t) { for(U=pre[u]; v!=s; V=u,u=pre[u]) {edge[cur[u]].cap-=k; edge[cur[u]^1].cap+=k; } m

Transferred from http://blog.csdn.net/regina8023/article/details/45815023 It 's written in front . Network flow with upper and lower bounds is limited by the flow of the edge, which must be within the range of [Down,up]. In fact, the ordinary network flow is a special network with the upper and lower bounds of the flow, but each side of the flow limit to [0,ca

lines. Each block describes one network. The first line of each block there is the number of places N by one space. Each block ends with a line containing just 0. The last block had only one line with N = 0;OutputThe output contains for each block except the last in the input file one line containing the number of critical places.Sample Input55 1 2) 3 4062 1 35 4 6 200Sample Output12HintYou need to determine the end of one line. In order to make it's

initial network.The next line contains a single integerQ(1≤Q≤1,000), which is the number of the new links the administrator plans to add to the network one by one.TheI-th Line of the followingQLines contains-integerAandB(1≤A≠B≤N), which is theI-th added new link connecting computerAandB.The last test was followed by a line containing the zeros.OutputFor each test case, print a line containing the test case number (beginning with 1) and Q lines, the i

graph theory algorithm-Network maximum flow template "Ek;dinic"ek TemplatesIncrease traffic by finding the minimum residue in the augmented residual network every timeconst int Inf=1e9;int n,m,s,t;struct node{int v,cap;}; vectorDinic templatesConstruct hierarchy + Block flow augmentationconst int Inf=1e9;int n,m;int s,t;int tot=1;struct node{int v,f,nxt;} E[10

loopExercisesEuler Circuit of network flow judgment mixed RoadThe key is to turn the graph into a map and then judgeIt is easy to know if it is the Euler loop, then all points of the in, equal to out of the degree outNo forward Edge (U,V), a forward edge (u,v), or a forward edge (V,u), in[i]-out[i] can be found unchangedThen we can assume that the non-u,v (u,v) becomes the forward edgeStatistics of all poi

Test instructions: Multiple sets of data, and the last 0/1 indicates 0 non-direction 1.Ask if there is a Euro-pull loop.Solving the puzzle : giving it any direction without a direction . First, the point-in degree of the Euler loop = the degree of exit.And then found that each non-directional side if the change direction, the original entry point of +1, out of 1, out of the point.Then we might as well be different in degrees and out of the point with the Yuanhui in one of the edges, the capacity

from all over the world who specialize in one type of aircraft, each of which has two drivers and requires a pilot and a co-pilot. For a variety of reasons, such as co-ordination, some pilots cannot fly on the same plane and ask how to match the driver to make the most of the flight., assuming that there are 10 drivers, the V1,V2,...,V10 represents up to 10 drivers, of which V1,V2,V3,V4,V5 is the pilot, and V6,V7,V8,V9,V10 is the co-pilot. If a pilot and a co-pilot can fly on the same plane, on

Codeforces 512C Fox And Dinner odd And even graph network flow, codeforces512c Question link: Click the open link Question: Given n people and their weights. Let people sit on any round table so that any two adjacent human rights values are prime numbers, and a table contains at least three people. Output The number of tables and the number of persons sitting at each table (output any specific scheme) Ideas

. The first line of each block there is the number of placesN N lines contains the number of a place followed by the numbers of some places to which t Here are a direct line from the this place. These at the veryN lines completely describe the network, i.e., each direct connection of both places in the network I s contained at least in one row. All numbers on one line is separated by one space. Each block e

; Bque.push (TEM); Vis[s]=1; while(!Bque.empty ()) {tem=Bque.front (); Bque.pop (); Mark (Tem.npoi,1);//according to the original image of the edge mark cannot go inttmp=LEN,TMP2; while(tmp--) {TMP2=nque.front ();//these points are not coming.Nque.pop (); if(!VIS[TMP2]) {Len--; Hh.step=tem.step+1; Hh.npoi=TMP2; Bque.push (HH); ANS[TMP2]=Hh.step; } ElseNque.push (TMP2); if(!len)return; } Mark (Tem.npoi,0); } return;}intMain () {intt,n,m; intu,v,s; s