cti transportation

This year I am back to my old business, for the work encountered problems and new learning knowledge, I will be in a free time to record!Yesterday called me to send a computer, installed the system, sent to the customer there, the customer said that the computer is too slow, the request to the host exchange and this, I think not to change the host, no problem! Do not change do not know, a change on a lot of things then come, the customer said that I change the machine how do not have the golden

four integers n (1OutputIncludes an integer representing the minimum total cost. Total cost =n days the sum of the length of the transportation route +k* Change the number of shipping routes.Sample Input5 5 10 81 2 11 3 31 4 22 3 22 4 43 4 13 5 24 5 242 2 33 1 13 3 34 4 5Sample Output +HINTThe first three days walk 1-4-5, after two days walk 1-3-5, so the total cost is () *3+ (3+2) *2+10=32Source/* ***********************************************autho

Problem descriptionW Company has m warehouses and N retail stores. The first warehouse has an AI unit of goods, and the first J retail stores need BJ units of goods. Balance of goods supply and demand . The cost of transporting each unit of goods from the I warehouse to the J retail store is c[i,j]. Try to design a transport plan that transports all the goods in the warehouse to the retail store, so that the total transportation costs are minimized. P

Advanced Seminar on Excellence in goal and performance management in the transportation industry Outline(Two- Day Combat Edition)(If you want to disclose the following, or you need to make a professional adjustment of the outline, please contact me to confirm)"Course keywords" performance management cycle, work plan, MBO\KPI\BSC, corporate culture"Course duration" 2 days (estimated 6.5 hours per day)"Course object" business leaders, department manager

Advanced Seminar on Excellence in goal and performance management in the transportation industry(Two- Day Forum Edition)(If you want to disclose the following, or you need to make a professional adjustment of the outline, please contact me to confirm)"Course keywords" performance management cycle, work plan, MBO\KPI\BSC, corporate culture"Course duration" 2 days (estimated 6.5 hours per day)"Course object" business leaders, department managers, other

Advanced Seminar on Excellence in goal and performance management in the transportation industry Outline( One Day Forum edition)(If you want to disclose the following, or you need to make a professional adjustment of the outline, please contact me to confirm)"Course keywords" performance management cycle, work plan, MBO\KPI\BSC, corporate culture"Course duration" 1 days (estimated 6.5 hours per day)"Course object" business leaders, department managers

DescriptionThe logistics company is going to ship a batch of cargo from Wharf A to Pier B. Due to the large volume of goods, it takes n days to complete the shipment. In the course of cargo transport, several terminals are generally diverted. Logistics companies typically design a fixed transport route to carry out strict management and tracking of the entire transport process. Due to the existence of various factors, sometimes a pier will not be able to load and unload goods. At this time, the

[CODEFORCES724E] Goods TransportationQuestion DescriptionThere is N cities located along the one-way road. Cities is numbered from 1 to n in the direction of the road. The I -th City had produced p i units of goods. No more Than s i units of goods can Sold in The I -th city. For the pair of cities I and J such that 1?≤? I? j. ≤? n You can no further than once transport no more than C units of goods by the city C14>i to the city J.Note that goods can is transported from

Rogu p31_lca, HDU2586, rogu P1967 freight car transportation, multiplication lca, tree multiplication, p3w.lcap1967 Multiply lca board loogu P3379 1 #include How far away? HDU-2586 (two points on the tree) The method is to find the distance from dis [I] to the root node, then two points, the distance between B is $ dis [a] + dis [B]-2 * dis [lca (a, B)] $ Error notes: Line 2 is written as anc [x] [I] = anc [fa] [I-1]; 2. 18 rows missing 1 #include L

lines is a description of the route, including three integers, which in turn represent the two port number of the route connection and the route length (>0). Pier A is numbered 1 and Pier B is M. The transportation cost per unit length is 1. The route is bidirectional. Then the next line is an integer d, followed by a row of D for each row is three integers P (1Output format:Includes an integer representing the minimum total cost. Total cost =n days

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

; - } - 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 for(inti =2; I ){ the if(Flag[i] Dis[i] min (Dis[maxi], street[maxi][i])) { +Dis[i] =min (Dis[maxi], street[maxi][i]); -

+-+-+.+-+-+|...| .....| +-+.+-+-+-+..| .......| 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];CharC;queueQ;voidpriintXinty) { intk,p; for(;;) {k=cm[x][y];p =0; if(k==0) printf ("E");Else if(k==1) printf ("S");Else if(k==2) printf ("W");Elseprintf ("N"); while(cm[x][y]==k) {x+

[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]; - } - A voidMax_flow (intXinty) + { the inta,sum=0, h=0; - BOOLok=0; $ while(a=BFS (x, y)) the { the intnow=y;sum+=a*Dis[y]; theh+=A; the while(now!=x) - { int[pre[now]].f-=A; thet[t[pre[now]].o].f+=A; thenow=t[pre[now]].x; About } the if(h>=

contains the number of scenarios (city plans). For each city the number N of the street crossings (1 OutputThe output for every scenario begins with a line containing "scenario #i:", where I am the number of the scenario starting at 1. Then print a, containing the maximum allowed weight that Hugo can transport to the customer. Terminate the output for the scenario with a blank line.Sample Input13 31 2 31 3 42 3 5Sample OutputScenario #1:4 Each road has a limited weightThe maximum number of good

(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)); } intD; scanf ("%d",D); while(d--) { intP, A, B; scanf ("%d%d%d", p, a, b); for(inti = A; I ) Days[i].push

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

];i;i=Next[i]) { the intPur=Go[i]; * if(flow[i]dis[pur]+1==Dis[x]) { $ intSave=dfs (Pur,std::min (flow[i],f-sum));Panax Notoginsengflow[i]-=Save; -flow[op[i]]+=Save; thesum+=Save; + if(dis[s]>=nodes| | F==sum)returnsum; A } the if(Flow[i]) mn=std::min (Mn,dis[pur]); + } - if(sum==0){ $cnt[dis[x]]--; $ if(cnt[dis[x]]==0){ -dis[s]=nodes; -}Else{ thedis[x]=mn+1; -cnt[dis[x]]++;Wuyi } the } - returnsum; Wu

]; at BOOLSPFA () { - for(intI=0; ii) { -D[i]=inf; inque[i]=0; - } -d[vs]=0; inque[vs]=1; -queueint>que; in Que.push (VS); - while(!Que.empty ()) { to intu=Que.front (); Que.pop (); + for(intI=head[u]; i!=-1; I=Edge[i].next) { - intv=edge[i].v; the if(Edge[i].cap d[v]>d[u]+edge[i].cost) { *d[v]=d[u]+Edge[i].cost; $pre[v]=i;Panax Notoginseng if(!Inque[v]) { -inque[v]=1; the Que.push (v); + } A } th