j j transportation

SPFA water problem. Record the maximum load capacity to this node. Because of a small error and WA a night, sad, t_t ...#include #include#include#include#include#includeusing namespacestd;Const intmaxn= ++Ten;structedge{int from, To,w;} e[1000010];intDis[maxn],vis[maxn];vectorG[MAXN];intN,m,tot;voidSPFA () {Queueint>p; memset (DIS,0,sizeof(DIS)); memset (Vis,0,sizeof(VIS)); vis[1]=1; Q.push (1); while(!Q.empty ()) { intH=q.front (); Q.pop (); vis[h]=0; for(intI=0; I) {Edgee=G[h][i]; if

com.baidu.mapapi.search.busline.BusLineSearchOption; Import Com.baidu.mapapi.search.busline.ongetbuslinesearchresultlistener;import Com.baidu.mapapi.search.core.cityinfo;import Com.baidu.mapapi.search.core.poiinfo;import Com.baidu.mapapi.search.core.searchresult;import Com.baidu.mapapi.search.poi.OnGetPoiSearchResultListener; Import Com.baidu.mapapi.search.poi.poicitysearchoption;import Com.baidu.mapapi.search.poi.poidetailresult;import Com.baidu.mapapi.search.poi.poiresult;import com.baidu.map

and K. (1 Outputoutput one, indicating the minimum cost. If It is impossible to transport all the K units of goods, output-1.Sample Input2 1 21 2 1 22 1 21 2 1 12 2 21 2 1 21 2 2 2Sample Output4-13Test instructionsThere are n cities and m one-way routes. For each line, there are four messages: The starting point A, the end point B, the cost factor C, the capacity D (1Analytical:This question is to use the cost flow to write, but the premise of the cost flow is proportional to the cost and flow,

Huang Feihong said: "Mr. Li, the winner is the king and the loser is the enemy. No one knows who won the fight if there is no skylight to block the gun. Now, the gold medal is in my hand, not my victory. In order to show the power of our people, we have lost so many deaths and injuries. According to the eyes of the minor, we do not only need to train and defend against external enemies. The most important thing is to broaden the wisdom of the people and unite the wisdom of the people. Can a bran

", a, b, len); D[A][B] = D[b][a] = len; }}voidPrim () {intK for(inti =1; I 1][i]; } vis[1] =1; for(inti =0; I 1; i++) {intL =-2; for(intj =1; J if(!vis[j] dis[j] > L) {L = dis[j]; K = J; } }if(Ans > L) ans = l;if(k = = N) Break; VIS[K] =1; for(intj =1; J if(!vis[j] d[j][k] > Dis[j]) {Dis[j] = d[j][k]; } } }printf("%d\n\n", ans);}intMain () {intT, Case =1;scanf("%d", t); while(t--) {printf("Scenario #%d:\n", case++); Init ();

Magnetic stripe card to change the chip card charges you can come to understand, the following will be one by one to introduce investment/CITIC/Minsheng/transportation/Bank of China for IC card fees, and some banks for IC card is free, and some is to be charged oh ~ together to understand it. --Bank of China Maximum of 5 yuan March 31 before the processing of magnetic stripe card change chip card does not charge. The new card charges 5 yuan. 6 Star

"CF 724E" Goods Transportation (min cut +DP) Main topic:n factories, each factory has a production pi p_i and a maximum sales Si s_i. Small numbered factories can deliver goods to large numbered factories, with a maximum of C units per plant. The routing sequence is arbitrary. Ask for the final sales of all factories. Can think of the maximum flow. The source points with each factory an edge, the Flow pi p_i, each factory and meeting point an edge, th

As we all know, network data transmission between different hosts is mainly completed through the TCP/IP network protocol. This is true for enterprise LAN data transmission and Internet data transmission. However, it is hard to figure out that no security was provided during the design of the TCP/IP protocol. That is to say, the TCP/IP protocol alone cannot guarantee the secure and stable transmission of data in the network. Therefore, the security of data in the network depends on high-level ap

Oracle Tablespace Transportation premise: the user must have SYSDBA system permissions for Tablespace transmission. The moved Tablespace is a self-contained Tablespace and should not depend on the existence of objects of external Tablespace objects. Determine whether to include the data. You can use the TRANSPORT_SET_CHECK process in the system package DBMS_TTS to check whether the data is transmitted to the table space OLTP. For example, you can use

; -memset (Street,-1,sizeof(street)); +memset (Flag,true,sizeof(flag)); A for(inti =0; I ){ atCin >> x >> y >>temp; -Street[x][y] = street[y][x] =temp; - } - for(inti =2; I ){ -Dis[i] = street[1][i]; - } in - for(intK =1; K ){ to intMaxx =-1, Maxi =1; + for(inti =2; I ){ - if(Flag[i] dis[i] >Maxx) { theMaxx =Dis[i]; *Maxi =i; $ }Panax Notoginseng } - theFlag[maxi] =false; + A

line contains a direction letters and a number of blocks to tour in that direction. Each row has a letter representing the direction and a Numbers at several intersections indicate.Sample Input3 6+-+-+.+-+-+|...| .....| +-+.+-+-+-+..| .......| s-+-+-+. e-+Sample OutputE 1N 1W 1N 1E 2S 1E 3S 1W 1Direct BFS, then record the path on the line ...#include #include#includeusing namespacestd;structna{intx, y;};Const intfx[4]={0,1,0,-1},fy[4]={1,0,-1,0};intn,m;Charmap[Bayi][Bayi];intcm[Bayi][Bayi];Char

]=0; flow[f1]=inf;inq[f1]=1; +dis[f1]=0; Q.push (F1); A while(!q.empty ()) the { + intx=Q.front (), y,i; - for(I=first[x];i;i=t[i].next)if(t[i].f>0) $ { $y=t[i].y; - if(dis[y]>dis[x]+t[i].c) - { thepre[y]=i; -dis[y]=dis[x]+t[i].c;Wuyiflow[y]=mymin (t[i].f,flow[x]); the if(!inq[y]) {Q.push (y); inq[y]=1;} - } Wu } -Q.pop (); inq[x]=0; About } $ if(pre[f2]==-1)return 0; - returnFLOW[F2]; - } -

) { intk=-1; intmin=-1; for(intI=1; i) if(!vis[i]dis[i]>Min) {Min=Dis[i]; K=i; } if(k==-1) Break; VIS[K]=true; for(intI=1; i) if(!vis[i]dis[i]min (dis[k],map[i][k])) {Dis[i]=min (dis[k],map[i][k]); } }}intMain () {intT; scanf ("%d",t); intCnt=0; while(t--) {CNT++; intm; memset (Vis,false,sizeof(VIS)); scanf ("%d%d",n,m); /*for (int i=1;i*/memset (Map,0,sizeof(map)); intu,v,w; for(intI=1; i) {scanf ("%d%d%d",u,v,W); MAP[U][V]=map[v][u]=W; }

; Que.push (v); } } } } returnd[m];}intCost (intIintj) {memset (used,0,sizeof(used)); for(intK = i; K ) for(intt =0; T ) Used[days[k][t]]=1; inttmp=SPFA (); if(Tmp==inf)returnINF; return(J-i +1) *tmp;}intMain () {//freopen ("1003.in", "R", stdin); //freopen ("1003.out", "w", stdout);scanf"%d%d%d%d", n, m, k, E); for(inti =0; i ) { intu, V, c; scanf ("%d%d%d", u, v, B); G[u].push_back (Edge (V, c)); G[v].push_back (Edge (U, c)); }

http://poj.org/problem?id=1797DescriptionBackgroundHugo Heavy is happy. After the breakdown of the Cargolifter project he can now expand business. But he needs a clever mans tells him whether there really is a-a-a-from-the-place his customer have build his giant stee L Crane to the place where it's needed on which all streets can carry the weight.Fortunately he already have a plan of the city with all streets and bridges and all the allowed weights. Unfortunately he have no idea how to find the

[j+1][E[I].V],1), Add (id[j][e[i].v],id[j+1][E[I].U],1); the intans=0; About while(Dis[s]DFS (s,inf); the returnans==J; the } the intMain () { + while(SCANF ("%d", n)! =EOF) { -M=read (); K=read (); Ss=read (); tt=read (); the for(intI=1; i)BayiE[i].u=read (), e[i].v=read (); the intL=1, r=n+k+2, ans=0; the while(lR) { - intMid= (L+R)/2; - if(Check (mid)) r=mid-1, ans=mid; the ElseL=mid+1; the } the

+=flow; -cost+=Edge[pre[u]].cost; A } +res+=cost*flow; the } - returnRes; $ } the the intMain () { the intn,m,k; the while(~SCANF ("%d%d%d",n,m,k)) { -vs=0; Vt=n; nv=vt+1; Ne=0; inmemset (head,-1,sizeof(head)); theAddedge (VS,1K0); the About intu,v,a,c; the while(m--){ thescanf"%d%d%d%d",u,v,a,c); the for(intI=1; ii) { +Addedge (U,v,1, A * (2*i-1)); - } the }Bayi intres=MCMF (); the if(Mxflow!=k) puts ("

--) { intn,m; CIN>>n>>m; intA,b,mid; CIN>>a>>b>>mid; Memset (Mat,0,sizeof(MAT)); for(intI=0; i) { intu,v; CIN>>u>>v; MAT[U][V]= Mat[v][u] =1; } //then you build the network streamMxf.init (0,0,2*n+2); for(intI=1; i) { if(i!=mid) Mxf.add_edge (2(In2*i+1,1); for(intj=i+1; j) { if(Mat[i][j] = =0)Continue; if(j!=mid) Mxf.add_edge (2*i+1,2*j,Ten); if(i!=mid) Mxf.add_edge (2*j+1,2(InTen); }} Mxf.s=2*mid+1; MXF.T=2*a+1;

the cost of rebuilding these T to S bidirectional path, then the collection of arbitrary points constitutes the S set also must not meet the above conditions ~ ~ ~ If the existence of a set satisfies the above conditions, there must be a single point to meet the above conditions ~~~~So we actually just use the enumeration point and then judge it. Orz ...Code#include Copyright NOTICE: This article for Bo Master original article, without Bo Master permission not reproduced. HDU 4940 Destroy

Test instructionsThere are N and cities and M roads, each city has the amount of gold and gold collected, and now it is time to collect all the gold, the shortest side of the pass.Analysis:binary + max stream or combined equivalence class with and check set.//poj 3228//sep9#include POJ 3228 Gold Transportation and collection