network graph software

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

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

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

ISAP algorithmISAP (Improved shortest augument Path) algorithm is an improved version of the SAP algorithm, if the efficiency requirements are high, the algorithm can be used.(1) Overview: The algorithm is based on the fact that the shortest distance from any node to a sink point (in a residual network) will not decrease after each augmentation. In this way, we can use the d[i[to represent the lower bound of the distance from node I to the sink point.

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

. 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