broadwell 1150

() - { - intres =0; -memset (linker,-1,sizeof(linker)); - for(intU =0; u ) in { -memset (Used,false,sizeof(used)); to if(Dfs (U)) res++; + } - returnRes; the } * intMain () $ {Panax Notoginseng inti,j,k; - //freopen ("1.in", "R", stdin); the while(SCANF ("%d", un)! =EOF) + { A if(un==0) Break; thescanf"%d%d",vn,k); + intp,q; -Memset (G,0,sizeof(g)); $ while(k--) $ { -scanf"%d%d%d",i,p,q); - if(p>0q>0) g[p

needs to be set to mode XI. If it runs on machine B, machine A needs to be set to mode Yi. Tasks on each machine can be executed in any order, but each machine needs to be restarted once for each conversion mode. Please arrange a machine for each task and arrange the order as little as possible. Minimum vertex overwrite of a bipartite graph = maximum number of matching This is the minimum vertex overwrite. Each task creates an edge. The minimum vertices overwrite means that the mini

)returnVal > a.val;returnx BOOL operator== (ConstABCD a)Const{returnx==a.x Val==a.val; }};namespaceheap{priority_queuevoidInsert (ABCD x) {Heap.push (x); }voidDelete (ABCD x) {Del_mark.push (x); }voidPop () { while(Del_mark.size () heap.top () ==del_mark.top ()) Heap.pop (), Del_mark.pop (); Heap.pop (); } ABCD Top () { while(Del_mark.size () heap.top () ==del_mark.top ()) Heap.pop (), Del_mark.pop ();returnHeap.top (); }}intN,K,ANS,A[M],VAL[M];namespaceunion_find_set{intFA[M],RANK[M],L[M]

clear the manager has strated the possible cases and impossible cases of simultaneous moving. For each room, at most one table will be either moved in or moved out. now, the manager seeks out a method to minimize the time to move all the tables. your job is to write a program to solve the manager's problem. Inputthe input consists of T test cases. the number of test cases) (T is given in the first line of the input. each test case begins with a line containing an integer N, 1 Outputthe ou

of data range and comment hint is the triangle with three vertices (-1,-1), (3, 1), (1, 3) respectively.The ropes enclose the following points:(-1,-1 ), (1, 3)10 in total.The answer may be out of the longint range. int64 or long is required. Source catfish pick theorem s = I + (A/2)-1; S indicates the area, I indicates the interior vertices of the polygon, and a indicates the edges; A is calculated using GCD. Pay attention to the case of 0. This question is a question with a large amount of da

machine can be executed in any order, but each machine will need to be restarted every time the mode is converted. Please properly arrange a machine for each task and arrange the order reasonably, so that the machine restarts as few times as possible. Analysis:minimum vertex cover number of a binary graph = maximum number of matchesthe point is to find the minimum number of vertex coverage. each task establishes an edge. The minimum point overlay is the least points that can be connected to all

Deepened my understanding of the minimum point coverage.Connect the two machines of each task, the problem is to select the least points to cover all the lines, that is, the minimum point coverage1#include 2#include 3#include 4#include 5 using namespacestd;6 Const intmaxn= the;7vectorint>map[ the];8 intLINK[MAXN],VIS[MAXN];9 intn,m,k;Ten BOOLDfsintt) One { A intI,x,size=map[t].size (); - for(i=0; i) - { thex=Map[t][i]; - if(!Vis[x]) - { -vis[x]=1; + if(li

][ the],vis[ the];8 9 BOOLFindintu)Ten { One for(intv=1; v) A if(map[u][v]!Vis[v]) - { -vis[v]=1; the if(!match[v]| |find (Match[v])) - { -match[v]=u; - return true; + } - } + return false; A } at - intMain () - { - for(intans=0; scanf ("%d", n) n;ans=0) - { -scanf"%d%d",m,t); in for(intx,y,z;t--;map[y][z]=1) -scanf"%d%d%d",x,y,z); to for(intI=1; i) + { - if(Find (

mode_0.Each task has a corresponding operating mode, (I, X, y) represents the operating mode mode_x, mode_y on the A B machine that corresponds to the I task. the tasks on each machine can be executed in any order, but each machine will need to be restarted every time the mode is converted. Please properly arrange a machine for each task and arrange the order reasonably, so that the machine restarts as few times as possible. and ask for its value. Ideas:Minimum point overlay (maximum number of

time you change the status. Ask how many times to restart these tasks. (Task completion is not sequential)Problem-solving ideas: two-dimensional diagram, the N of a machine as the left, B machine's M state as the right, (Xi,yi) constitute an edge, Hungary to find the maximum number of matches can be.#include #include#includeusing namespacestd;Const intmaxn= the;intLINE[MAXN][MAXN];BOOLUSED[MAXN];intN,M,K,MATCH[MAXN];BOOLDfsintx) { for(intI=1; i//tasks with a status of 0 are done directly at

means it can be processed either A at mode_x, or in machine B at mode_y.Obviously, to accomplish all the jobs, we need to change the machine's working mode from time to time, but unfortunately, The machine's working mode can only is changed by restarting it manually. By changing the sequence of the jobs and assigning each job to a suitable machine, please write a program to minimize the Times of restarting machines.Inputthe input file for this program consists of several configurations. The fir

Group A n person B group M person//MAX How many people match each person just once # includeHDU 1150 Machine Schedule (classic binary match)

][i]; A if(!Vis[v]) at { -vis[v]=true; - if(link[v]==-1||DFS (Link[v])) - { -link[v]=u; - return true; in } - } to } + return false; - } the intMain () * { $ //freopen ("A.txt", "R", stdin);Panax Notoginseng inta,b,c; - while(~SCANF ("%d", n) N) the { +scanf"%d%d",m,k); A for(intI=0; i) g[i].clear (); the for(intI=0; i) + { -scanf"%d%d%d",a,b,c); $ if(b>0c>0)

1150: [CTSC2007] Data backups backup Time Limit:10 Sec Memory limit:162 MBsubmit:797 solved:333[Submit] [Status] Description Input The first line entered contains integers n and K, where n (2≤n≤100 000) represents the number of office buildings, and K (1≤K≤N/2) represents the number of network cables available. The next n rows contain only one integer (0≤s≤1000 000 000), representing the distance from each office building to the beginning of the stre

The method used here is very violent, that is, the direct BFS; Only 1150, just as practice under BFS//source code of submission 804105, Zhongshan University Online Judge System #include

For typical BFs, the first one is self-written without any optimization. It requires 8 ^ 8 space storage status. Later, I came to the others' problem-solving report and learned how to expand Kanto and compress them in full order. I only needed 8!

Machine Schedule Time Limit: 2000/1000 MS (Java/others) memory limit: 65536/32768 K (Java/Others)Total submission (s): 5817 accepted submission (s): 2932Problem descriptionas we all know, machine scheduling is a very classical problem in computer

There is a sequence of work to complete. Now there are two machines A, B, and machine A in n modes, and machine B in M modes, each task can be completed in the mode on machine A, or in the B mode of machine B. You need to restart the machine when

Machine Schedule Time Limit: 2000/1000 MS (Java/others) memory limit: 65536/32768 K (Java/Others)Total submission (s): 5371 accepted submission (s): 2658 Problem descriptionas we all know, machine scheduling is a very classical problem in computer

/* If you want to determine whether to use DFS or BFs, the DFS may produce incorrect results. For example, if you use aaab to implement bfsconstraintstime limit: 1 secs, memory limit: 32 MB, the special judgedescription magic board consists of eight