mindstorms nxt

Conclusion: Each time the character is dropped to the outermost optimal, with a tree-like array of statistical answers, the characters placed on the outermost can be considered to disappear, directly in the tree-like array to delete the good.Perceptual understanding is that throwing characters into the middle increases the number of other characters moving, but it's not right to throw them outside.#include #include#include#include#definell Long Longusing namespacestd;Const intmaxn=500010;intn, x

Facet links/*code correctness is not guaranteed. in the DFS sequence do mo teamwhen a point is not another point of the LCA, it is necessary to add their LCA contribution*/#include #include #include #include #define GC () GetChar ()Const intn=4e5+5, m=1e5+5;intn,m,size,enum,h[n],nxt[n1],to[n1],dep[n],fa[n],top[n],son[n],sz[n];intnow,ans[m],seq[n1],in[n],out[n],id,a[n],tm[n];BOOLVis[n];structques{intL,r,lca,id,x,y;BOOL operatorConstQues a)Const{returnl

[Topic link]http://poj.org/problem?id=3694AlgorithmFirst, we use the Tarjan algorithm to find all the side of the dual-link components, and then, the image of the contraction pointIf the added edge (x, y) is in the same double-unicom component, the answer is the same, otherwise, the edges on the path to Belong[x]-belong[y] are marked, and the process can be accelerated by using the setCode#include #include#include#include#include#include#include#include#include#include#include#include#include#in

discards the fragment and returns the RST. (This is the mechanism of TCP processing without port sniffing) if the protocol control block exists, but the state is closed, the server does not invoke connect () or listen (). The fragment is discarded, but the RST is not returned. The client will attempt to re-establish the connection request. Create a new socket: when the socket in the listening state receives the segment, a child socket is created, along with socket{},tcpcb{} and pub{. If an err

$ (base64, W + 1, 1) Else Mimeencode = "" End Function Private Function Mimedecode (A as String) as Integer If Len (a) = 0 Then Mimedecode = -1:exit Function Mimedecode = InStr (base64, a)-1 End Function Public Function Encode (ByVal Inp as String, ByVal e as Long, ByVal n A S Long) as String Dim S as String s = "" m = INP If m = "" Then Exit Function s = mult (CLng (ASC (Mid (M, 1, 1)), E, N) For i = 2 to Len (m) s = S "+" Mult (CLng (ASC (Mid (M, I, 1)), E, N) Next I Encode = Base64_encod

back to FJ's farm to visit her friends. Bessie liked the roadside scenery so much that she did not want to end her journey so quickly, so every time she returned to the farm, she chose the second shortest path, rather than the shortest one, as we used to. Bessie's Village has R (1Input/output formatInput format: Line 1:two space-separated integers:n and R Lines 2..r+1:each Line contains three space-separated integers:a, B, and D this describe A road that connects Intersecti ONS A and B and has

Topic links\ (description\)Each point has a cost Si and value pi, which requires selecting some connected blocks with a root, the total size is k, making \ (\frac{∑pi}{∑si}\) the largest\ (solution\)01 Fractional plan, then DP, set F[I][J] indicates the maximum weight of the I subtree selected J and, direct violent backpack transfer can beWhen enumerating the number of child nodes selected, assume X has 1.2.3.4 four child nodes with a complexity of \ (1*sz[1]+sz[1]*sz[2]+ (sz[1]+sz[2)) *sz[3]+ (

", X) One #definePL (x) printf ("%lld\n", X) A #defineRep (I, A, n) for (int i=a;i - #definePer (i, a, n) for (int i=n-1;i>=a;i--) - #defineFi first the #defineSe Second - using namespacestd; -typedef pairint,int>PII; - Const intN = 1e5 +5; + Const intMoD = 1e9 +7; - Const intMOD =998244353; + ConstDB PI = ACOs (-1.0); A ConstDB EPS = 1e-Ten; at Const intINF =0x3f3f3f3f; - Constll INF =0x3fffffffffffffff; - - intn,m,cnt; - BOOLmp[205][205]; - intDu[n],head[n]; in inta[205]; - structp{intTO,

3198: [sdoi2013]springtime limit:40 Sec Memory limit:256 MBsubmit:1143 solved:366[Submit] [Status] [Discuss] DescriptionInputOutputSample Input3 31 2 3 4 5 61 2 3 0 0 00 0 0 4 5 6 Sample Output2 HINTDragonite Correcting Data SourceHash simple , but hash is a bit troublesome.Ans= at least the logarithm of K *c (K,K)-at least k+1 logarithmic *c (k+1,k) + at least k+2 logarithmic *c (k+2,k) ...For this C (i+k,k), my understanding of this logarithm was computed C (k+i,k) timesHash judg

[Topic link]http://poj.org/problem?id=3417AlgorithmDifferential on treeCode#include #include#include#include#include#include#include#include#include#include#include#include#include#include#include#include#include#include#include#include#include#include#include#include#include#includeSet>#include#include#include#includestring>#include#include#include#include#include#includeusing namespacestd;#defineMAXN 100010#defineMaxlog 20structedge{intTO,NXT;} E[MA

maintain, so consider the insertion of the point of destruction,when you insert point now[i], it means that now[i + 1].....now[k] has been inserted (not yet destroyed), and Now[1]........now[i-1]has been inserted (destroyed), so you can use and check set, to maintain the unicom block. #include using namespacestd;Const intMAXN =400005;intN, M, Fa[maxn], K, VIS[MAXN], ANS[MAXN];intHEAD[MAXN], cnt =1, tot =0, NOW[MAXN];structnode{intV, NXT;} G[MAXN];voi

#include using namespacestd;Const intMAXN = 3e5 +Ten;structedge{intV, NXT;} EDGE[MAXN];CharCH[MAXN];intDEG[MAXN];intdp[maxn][ -];intHEAD[MAXN], CNT;intN, M;inlinevoidinit () { for(intI=0; i) {Head[i]= -1, deg[i] =0; for(intj=0; j -; J + +) Dp[i][j]=0; } CNT=0;} InlinevoidAddedge (intFrom,intTo ) {EDGE[CNT].V=to ; EDGE[CNT].NXT=Head[from]; Head[from]= cnt++;}#defineDebug 0intMainvoid){#ifDebugFreopen ("In.t

++;} InlineintRead () {CharC=NC ();intx=0, f=1; while(c'0'|| C>'9'){if(c=='-') f=-1; c=NC ();} while(c>='0'c'9') {x=x*Ten+c-'0'; c=NC ();} returnx*F;}intn,m,s,t;intH[MAXN];//the height of each nodeintF[MAXN];//traffic that can flow from each nodeintGAP[MAXN];//the number of each heightstructnode{intU,V,FLOW,NXT;} EDGE[MAXN];intHEAD[MAXN];intnum=0;//Note here num must start from 0InlinevoidAdd_edge (intXintYintz) {edge[num].u=x; EDGE[NUM].V=y; Edge[n

not appear a cow is all the cattle like.Note that there was an error in the ID record of my connected component at the beginning because my first point was in the main function, causing some points to not be recorded. Later affected by the inertial thinking of the graph, and the Judge PA! The condition of the = v Plus, this is the direction of the graph, is definitely not to add this judgment! The specific code is as follows:#include #include#include#includeusing namespacestd;#defineMAXN 10010s

/*the indentation plus the longest road should use topological sequences don't want to write SPFA run for a minute.*/#include#include#include#include#include#defineM 500200#definell Long Longusing namespacestd;intBe[m], ed[m], to[m], nxt[m], DFT, Head[m], CNT, belong[m], dfn[m], low[m], dis[m], ver[m], ver2[m], st[m], top, tot;BOOLVis[m], isj[m], zz[m];intRead () {intNM =0, F =1; Charc =GetChar (); for(;!isdigit (c); c = GetChar ())if(c = ='-') F =-1

the outside world, and the column and S edge capacity of the first person is 1.Then ISAP run the network stream.This pit point: pay attention to the difference in the output of the single complex output　　　　A long absence of 1A#include #include#include#include#include#include#include#include#include#includestring>#include#includeSet>#include#include#defineLson Rt#defineRson Rt#defineFi first#defineSe Second#definePing (x, y) ((x-y) * (x-y))#defineMST (x, y) memset (x,y,sizeof (x))#defineMCP (x,

manager and dispatch a third and fourth Ninja, the sum of the salaries is 4, not exceeding the total budget of 4. Since 2 ninjas were dispatched and the manager's leadership was 3,User satisfaction is 2, is the maximum number of user satisfaction can be obtained.Heuristic merging of TreapThe answer to each subtree U is l[u]*k (k is a small node in the U subtree and its weight is exactly less than or equal to m)And then use Treap to merge each subtree into a subtree to maintain the subtrees of t

(); returnw?-x:x;} ll f[2][n];intn,k,nxt[n][k],q[n],s[n],l,r,now=1, pre=0; inline ll Sqr (ll K) {returnk*K;} Inline DL Suan (intJintk) { if(S[j]==s[k])return-1e18; intI=Pre; return(F[I][K]-F[I][J]+SQR (S[j])-SQR (S[k])/(DL) (s[j]-s[k]);}intMain () {n=read (), k=read (); for(intI=1; i1]+read (); for(intj=1; j) { now^=1, pre^=1; l=r=0; for(intI=1; i){ while(L1]) ; intt=Q[l]; F[now][i]=f[pre][t]+ (LL) (S[i]-s[t]) *s[t];

Idea: Use the Tarjan algorithm to find the cut point, in the enumeration to remove each cut point can form the number of Unicom block.Note: Later I looked at the other code, found that my enumeration of the way to cut points is a rather stupid way, we can completely in the Tarjan process to find out the answer, introduce the discussion:If this cut point is the root node, in the Tarjan algorithm to find a few children node (Low[v] >= dfn[u]), he can cut a few unicom block, if the cut is a child n

#defineRson m+1,r,rt#defineRead Freopen ("In.txt", "R", stdin)#defineWrite Freopen ("OUT.txt", "w", stdout)#defineN 155intMp[n][n];intDis[n][n];intNxt[n][n];stackint>St;intFloydintN) { intAns =inf; for(intK =1; K k) { for(inti =1; I i) for(intj = i+1; J j) {if(Ans > Dis[i][j] + mp[k][i] +Mp[j][k]) {ans= Dis[i][j] + mp[k][i] +Mp[j][k]; while(!st.empty ()) St.pop (); for(intt = i; T! = J; t =Nxt[t][j]) St.push (t); St.push (j);