psn fee

; Res+=d*H[t]; for(intv=t;v!=s;v=Prevv[v]) {EDGEe=Es[preve[v]]; E.cap-=D; ES[PREVE[V]^1].cap+=D; } } returnRes;}voidclear_g () {EH=0; memset (Head,-1,sizeofhead);}Const intmaxn=1e4+ -;intm,n,k;intA[MAXN],W[MAXN];intLAST[MAXN];intTol;inthh;voidbuild () { for(intI=1; i) {tol+=W[a[i]]; if(Last[a[i]]) Addedge (last[a[i]],i-1,1,-W[a[i]]); Last[a[i]]=i; } for(intI=1; i1; i++) Addedge (i,i+1Inf0);}intMain () {Freopen ("/home/slyfc/cppfiles/in","R", stdin); //freopen ("/home/slyfc/cppfiles/ou

; ++tail; Q[tail]=s; B[s]=false; while (Head { ++head; int U=h[q[head]]; while (u!=0) { if (W[u]+dis[q[head]] { Dis[to[u]]=dis[q[head]]+w[u]; Pre[to[u]]=u; if (B[to[u]]) { B[to[u]]=false; ++tail; Q[tail]=to[u];}}U=next[u];}B[q[head]]=true;}if (dis[t]}void Get (){ int now=t; int mx=0x7fffffff; while (Now!=s) { Mx=min (Mx,flow[pre[now]); Now=from[pre[now]];}now=t;while (Now!=s){Ans+=mx*w[pre[now]];FLOW[PRE[NOW]]-=MX; FLOW[PRE[NOW]^1]+=MX;Now=from[pre[now]];}}int main (){Len=1; s=0;N=read (); M=

Click to open linkTest instructions: To give a graph without direction, ask you to delete any side, so that the graph is a Hamiltonian graph, if there are multiple, the output path on the ownership value and the smallest, no output noIdea: Starting today from more than 10 years of school began to brush questions, the enemy left us a little time, the problem after reading test instructions, looked at the sample, his yy a bit, wrote a pitch, AC, read the sample after I was thinking, because it is

, and finished.First of all the sum of the AI, and all the sum of the BI, if the two do not want to wait, the number is different before and after, it is impossible to do, direct pruningRemember to verify that the maximum stream is equal to the sum of all the digits of bi and, if not equal, that not all can be remitted to the meeting point, indicating that the state is not reachedOtherwise, the minimum cost is output.#include Copyright NOTICE: This article for Bo Master original article, withou

emotional small fluctuations, largehome to understand each other, because everyone is to a goal to work hard, there is any problem between everyone to say clearly OK. As the team leader, the task assignment shouldevenly, to inject a positive emotion into the whole team and make the whole project team combative. The team members should obey the arrangement of the group leader, do their own work,The project has any comments and suggestions, you can actively communicate with the leader through the

HDU 2686 MatrixTopic links3376 Matrix AgainTopic linksTest instructions: The two questions are the same, only the data range is not the same, is a matrix, from the upper left to the lower right corner from the lower right corner to go to the upper left corner to get the maximum valueIdea: To split the point. Build the map, then run the cost flow can, just HDU3376 this problem, the limit is 300W edge, and then the card time over 2333Code:#include Copyright notice: This article blog original arti

=s+1; the intu,v,w; -for (I,0, N) for (J,0, N) d[i][j]=INF; thefor (I,1, M) { thescanf"%d%d%d",u,v,W); theD[u][v]=d[v][u]=min (D[U][V],W);//Heavy Edge94 } theFor (K,0, n) for (I,0, N) for (J,0, N) the if(Kd[k][j],d[i][j]); theMc. Addedge (s,n+1K0);98for (I,1, N) { AboutMc. Addedge (s,i+n+1,1,0); -Mc. Addedge (I,t,1,0);101 }102for (I,0, n) for (j,i+1, N)103 if(D[i][j]!=inf) MC. Addedge (i+n+1J1, D[i][j]);104 LL cost; the MC. Mincost (s,t,cost);106printf"%lld", cost);10

1 Disablesceneredraw ()2 Try3 (4 For fe in geometry do5 (6 if Fe. Mesh.numfaces = = 0 Then7 ( 8 Delete fe;9 ) Else ()Ten ) One For i in Geometry do A ( - max create mode; - Select I the --subobjectlevel = 1 - if classof i==editable_poly Then - ( - i.editablepoly.deleteisoverts () + ) Else - ( + if classof I==editable_mesh Then A ( at meshop.deleteisoverts i; - ) Else () - ) - --subobjectlevel = 0 - ) - ) catch

The topic: Given the number of students and the number of study groups, students to participate in the group need to pay fees, each group will spend c[i]*cnt[i]^2. Each student can participate in a K-group, asking the most students to attend the minimum expenses incurred.Thought: If not count the next what ghost condition, see the picture is very obvious.S-> each student f:k,c:0Each student, each study group F:1,c:-f[i]Each Study group->t f:1,c:1 * c[i],3 * c[i],5 * c[i],7 * C[i],......The latte

entering that house. InputThere are one or more test cases in the input. each case starts with a line giving two integers n and m, where N is the number of rows of the map, and m is the number of columns. the rest of the input will be n lines describing the map. you may assume both N and m are between 2 and 100, aggressive. there will be the same number of 'H's and 'M' s on the map; and there will be at most 100 houses. input will terminate with 0 0 for N and M. OutputFor each test case, output

will be followed by a line containing zero. OutputFor each test case, output a single integer on a line by itself-the number of seconds you and your friend need between the time he leaves the jail cell and the time both of you board the train. (assume that you do not need to wait for the train-they leave every second .) if there is no solution, print "back to jail ". Sample Input Sample output 211 2 999331 3 102 1 203 2 509121 2 101 3 101 4 102 5 103 5 104 5 105 7 106 7 107

Although the question is for the maximum cost, we can convert it into the minimum cost. Use the difference between a number greater than the maximum value and each number as the cost value. After the final processing, I can do it. Some people simply multiply each value by-1, which is easier. I have been thinking about this question for a long time and asked some people. Finally, some knowledge is obtained. On the 214 page of the challenge Programming Competition. Point has a capacity limit, it m

This question is which course is selected at the current time. It is easy to know that the selection is related to the score and is a changed weight. Therefore, you can use the split method, split all the points from base to 100 into points. However, if you split the points like this, you must ensure the order when running the billing flow, which is troublesome .. Observe the formula and find that in the same course, the higher the score, the lower the weight, so this is monotonous. In this way,

// Minimum fee stream

POJ Training Plan 2195_Going Home (network stream/fee Stream) Solution report Question Portal Ideas: Create a bfs graph and run the billing flow. #include #include #include #include #define inf 0x3f3f3f3fusing namespace std;struct E { int v,cost,cap,next;} edge[100000];int head[1000],cnt,dis[1000],pre[1000],vis[1000],s,t,mmap[110][110],h,w,f[1100],cost,flow;struct N { int x,y,step;};void add(int u,int v,int cost,

HDU 4862 Jump minimum k path overwrite fee stream, hdu4862 Gg... Question: Matrix given n * m Select Q: The minimum cost of traversing the entire Graph If the traversal fails, the output is-1. # Include

Topic: Given a plane of some points, eat Mr. Bean from the origin, can only go right or upward, beg two eat Mr. Bean up to how many beansEach point is split into two, with a flow rate of 1, and a cost of 1 sides;If you can get to another point from one point, connect the out point of the previous point back to the point in the pointRun the cost stream. But it's obviously going to be tle.If I can go to J,j to K, then obviously need not even i->k this edge this is a pruningAfter adding this prunin

Redhat's Yum online update is chargeable, not registered and cannot be used, the workaround is to replace the Yum with CentOS, the following is the detailed process:1. Uninstall Redhat's original YumRpm-e ' Rpm-qa |grep yum '--nodeps2. Download the Yum installation package for CentOS to NetEase 163 image sourcewget http://mirrors.163.com/centos/6/os/x86_64/Packages/yum-3.2.29-60.el6.centos.noarch.rpmWgethttp://mirrors.163.com/centos/6/os/x86_64/Packages/yum-metadata-parser-1.1.2-16.el6.x86_64.rp

input parameters and returned values. ③ Reduce code duplication: abstraction can reduce a large amount of code. "generic" is a small part of abstraction, so "generic" must be learned. ④ The Only Way to object orientation: the three elements of object orientation are based on abstraction, while "generic" is a small part of abstraction. Therefore, "generic" is the only way to object orientation. ⑤ The beginning of architecture: to do a good job of architecture, object-oriented must be mastered, a

Idea: I didn't know much about this question at the beginning. Then I thought it was the biggest bare-cost stream. Just do it after creating a picture. However, the question is that binary matching is used to do this. Because the weighted matching method does not, the maximum stream with the minimum cost can be used directly. # Pragma comment (linker, "/Stack: 1024000000,1024000000 ") # include Maximum fee of wikioi 1028