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;void addside (int a,int b,int c) {Side2[top]= (Side2) {b,c,node[a]};node[a]=top++;Side2[top]= (Side2) {a,0,node[b]};node[b]=top++;}int getflow (int u,int f) {if (u==n+1)return F;int ans=0;for (int i=node[u];i!=-1;i=side2[i].next) {int to=side2[i].to,c=side2[i].val;if (dis[u]>dis[to]c) {int X=getflow (To,min (f-ans,c));Side2[i].val-=x;Side2[i^1].val+=x;Ans+=x;if (ans==f) return ans;}}gap[dis[u]]--;if (gap[dis[u]]==0) dis[0]=cnt+2;dis[u]++;gap[dis[u]]++;return ans;}bool OK (int x) {cnt=n+2;memset

, N; - Long LongSPFA () + { A Edge P, Pn; atP.E =1, P.W =0; -Queue Q; - Q.push (P); - - while( !q.empty ()) - { inP =Q.front (); - Q.pop (); toVIS[P.E] =false; + intLen =g[p.e].size (); - the for(intI=0; i) * { $Pn =G[p.e][i];Panax Notoginseng - if(DIST[PN.E] min (DIST[P.E],PN.W)) the { +DIST[PN.E] =min (DIST[P.E],PN.W); A the if(!VIS[PN.E]) + { -VIS[PN.E] =true; $ Q.push (Pn); $ } -

In fact, the retrospective is very simple is the violence + judgment conditions, the conditions of the establishment of continued violence, conditions are not set to return to the next violence, the difficulty is to find the conditions for the establishment,In my opinion, backtracking is the pruning of violence, to find out the condition of the establishment of the basic solution ...Ideas:This question should find out the current bus order and the number of people can get on the train,, I used n

; theMAXF =0; the while(SPFA ()) + { -MAXF + =A[t]; theCost + = a[t] *D[t];Bayi intU =T; the while(U! =s) the { -Edges[p[u]].flow + =A[t]; -edges[p[u]^1].flow-=A[t]; theU = edges[p[u]]. from; the } the } the returnCost ; - } the the intMain () the {94 while(SCANF ("%d%d%d", n, m, k) = =3) the { theInit (n +1); the 98s =0, t =N; AboutAddedge (s),1K0); - 101 while(m--)102 {103 intu, V, a, C;104scanf"%d%d%d%d",

using namespace STD;Const intMAXN =1010;intV, G, N, V[MAXN], G[MAXN], W[MAXN];intF[MAXN][MAXN];voidInit () {scanf("%d%d%d", v, g, n); for(inti =1; I scanf("%d%d%d", w[i], v[i], g[i]); }}voidWork () { for(inti =1; I for(intj = V; J >= V[i]; --J) { for(intK = G; K >= G[i]; --K) {F[j][k] = max (F[j][k], F[j-v[i]][k-g[i]] + w[i]); } } }printf("%d\n", F[v][g]);}intMain () {init (); Work ();return 0;} Copyright NOTICE: This article for Bo Master original article, without Bo Mast

, select "Open in Browser" in the "open document in Browser" item;Then, in InfoPath, set the display of the Ribbon or toolbar when the Web browser opens, and Hide tab buttons such as SubmitWhen not set, the Ribbon or toolbar, such as:Set method: "File"-"info"-"form Options"-"Web browser" in InfoPath, tick "commit" "Save" and "Save As" "Update" buttonAfter setting the effect asFinal Publish Form select Publish form to SharePoint libraryIn specify what you want to create live modifications, select

Link:http://poj.org/problem?id=1797Code:#include #include#include#includestring.h>#include#include#includeusing namespacestd;#defineN 1100#defineINF 0x3f3f3f3f3fintN, M, Dist[n], g[n][n], v[n];intDIST (intSintE) {dist[1]=0; v[1]=1; for(intI=1; i) Dist[i]= g[1][i]; for(intI=1; i) { intindex=-1, max=-1; for(intj=1; j) { if(v[j]==0 dist[j]>MAX) {Index= J, MAX =Dist[j]; }} V[index]=1; for(intj=1; j) { if(v[j]==0) { intTMP =m

GIS outsourcing of zhongke yanyuan --- transportation integrated Geographic Information Platform, yanyuan gisA gis portal that integrates maps, services, and transport. It is a GIS portal that integrates maps, services, and transport, without complex settings and deployment, you can quickly create interactive maps and applications and share them with other people in the company. To improve work efficiency. Platform: Flexible data management and shar

Add the code to world. save. [OBJECT]GUID = 0211B2BTYPE = 5ENTRY = 176495.MODEL = 3031XYZ = 2056.709473 237.530640 99.766960 4.839873GTYPE = 15GFLAGS = 40 This Zeppelin is grom' Gol Related transportation tools and items . Addgo 4170-orc platform elevator down. Addgo 4171-orc platform lift up. Addgo 20649-undervator). Addgo 20650-open/close the door of the undead underground city . Addgo from 176080 to 176085 is the subway . Addgo 164871The airships

Tags: bzoj short circuit dynamic plan zjoi2006 spfa Some docks are composed of several edges. In some cases, some docks need to be repaired and cannot be used during this period. If you change the route, you will have to spend a certain amount of time to find the minimum cost within the specified number of days. Idea: Use spfa for short circuit. The key is dynamic planning. I thought about this rule for a long time and finally found myself thinking complicated. I first thought about how to use

DescriptionA has n cities in China, ranging from 1 to n. There are m two-way roads between cities. Each road imposes weight limitations on vehicles. Now there are Q trucks carrying goods. drivers want to know that each truck can carry a maximum of multiple goods without exceeding the vehicle weight limit. Input FormatThe first line has two integers n, m separated by a space, indicating N cities and M roads in.Next, line M contains three integers x, y, and z in each row. Each integer is separated

that already exist is greater than MAXSIZE, need to open up space to store new cities and ++maxsize7. Manually add lines8. Insert Line: Because the line information is stored in the table node, you need to create a new table node and add a table corresponding to the starting city9. Read the line from the file10. Deleting a line11. Find the least cost path12. Find the minimum time pathCodeOn the GitHub link: https://github.com/bw98/National-Transport-AdvisoryNational

Topic Link: [POJ 1797]heavy TRANSPORTATION[SPFA] The analysis: For all paths from point 1 to N, the maximum size of the smallest capacity on the path. Ideas for solving problems: will be the shortest way of thinking change on the line. The Dis[i] Array records the smallest road capacity on the road from point 1 to I. Note Initialization of the DIS array. Personal experience: It was written because it was stuck in the initialization and tried several t

the Given an M-without-direction graph, you need to plan N-day transportation solutions. N days, every day from the S walk to T walk once, each point in some days can not pass. The cost is equal to the length of the path each day, and if the path chosen by the first day is different from the I-1 Day (i>1), it will take more than K to change the route price. Ask for the minimum total cost.(N Solving Brain hole problem, not too difficult. May at first

It's just a second answer, and then consider how to validate.Meet the Bi−ai≤mid b_i-a_i \le mid point on the tube, set the transmission point built in X,y X,y, the remaining points need to meet:| x−ai|+| Y−bi|≤mid | x-a_i|+| Y-b_i|\le midThis is

The bare maximum spanning tree +lca=. =, two templates just a little bit better =w= Type Rec=record ll,rr,len:longint; End var n,m,l:longint; I,j:longint; A:ARRAY[0..50010] of rec; F,D,LAST:ARRAY[0..10010] of Longint;

EndurerOriginal 1Version Code added to the first part of the homepage of the website:/-------- --------/ Code added to the central part:/-------- --------/ 1. hxxp: // G *** M *** 06. Z *** KD *** 5366cn/kkk/wm.htm Contains VBScript programs

A strongly connected and directed simple graph. Each edge has two values, D and B. Set S to a non-null real subset of the point set. Q: Is there sum (d (I, j) No idea at all during the competition... The maximum stream is still a whiteboard. Idea:

This question is about selecting the maximum load path from 1 to n and finding the maximum load. To obtain the maximum load, we need to find the minimum load that each route can carry. You can use Dijkstra algorithm to find Assume that the current

Want to see more problem-solving reports:Http://blog.csdn.net/wangjian8006/article/details/7870410Reprinted please indicate the source: http://blog.csdn.net/wangjian8006 There are n cities and M roads with a carrying capacity on each road. Now the