cod warfare

+1) lsum[root]+=lsum[root1|1]; if(Rsum[root]==r-mid) rsum[root]+=rsum[root1];}voidBuildintRootintLintR) { if(l==R) {Node[root]=lsum[root]=rsum[root]=1; return; } intMid=l+r>>1; Build (Lson); build (Rson); Pushup (Root,l,r,mid);}voidAddintRootintLintRintflag) { if(l==R) { if(flag==1) lsum[root]=rsum[root]=node[root]=1; Elselsum[root]=rsum[root]=node[root]=0; return; } intMid=l+r>>1; if(xmid) Add (Lson,flag); ElseAdd (Rson,flag); Pushup (Root,l,r,mid);}intQueryintRootintLintR

].len ()) A[r].ls= A[lson].ls +A[rson].ls; if(A[rson].rs = =A[rson].len ()) a[r].rs= A[rson].rs +a[lson].rs; A[r].s= max (a[lson].rs +a[rson].ls, Max (A[lson].ls, a[rson].rs));}voidBuildsegtree (intRintLintR) {A[r]. L= L, A[r]. R =R; A[r].ls= A[r].rs = A[r].s =A[r].len (); if(L = = R)return; Buildsegtree (Lson, L, A[r].mid ()); Buildsegtree (Rson, A[r].mid ()+1, R);}intQuery (intRintx) { if(A[r].s = =0) return 0; if(x //Judging if it's on the left returnA[r].ls; if(x > A[r].

intres=0;Bayi if(p10); the ElseRes+=doquery (Rson) + (p-m0); the returnRes; - } - } the the intQueryintp) { the This->p=p; the returnDoquery (1N1); - } the #undefInit the #undefLson the #undefRson94 }; the the intN, M; the intStk[maxa], top;98 intMain () { About #ifndef Online_judge -Freopen ("inch","R", stdin);101 //freopen ("Out", "w", stdout);102 #endif103 while(~SCANF ("%d%d", n, m)) {104 Segment

the first5~7Tin Hau "Qiqiao bleed" not only. Because of acute necrosis of the liver, spleen, lungs and kidneys, decomposition (decay). Patients continue to vomit the necrotic tissue from the mouth, and finally more because of extensive internal bleeding, brain damage and other causes of death. All acute necrosis (decay) of the internal organs of Ebola patients. It's scarier than the atomic bomb (for a moment). It is not impossible to use microorganisms (vicious viruses) to modernize wars. Scie

); while(!q.empty ()) {intU = Q.front (); Q.pop (); Vis[u] =false; for(inti = Head[u]; I! =-1; i = e[i].next) {if(i = = ID | | i = = ID +1)Continue;intv = e[i].to;if(D[v] > D[u] + e[i].dist) {D[v] = D[u] + e[i].dist;if(!vis[v]) {Q.push (v); VIS[V] =true; } } } }intS =0; for(inti =1; I returnS;}voidSolve () {intS =0; for(inti =1; I 0; SPFA1 (i); S + = Sum[i]; }intMax =-inf; for(inti =0; I intt =0;intx = node[i].x, y = node[i].y, id = node[i

"%lld%lld\n", first, second);//only 1 spaces the }98 About intMain () - {101Freopen ("Input.txt","R", stdin);102 intA, B, C;103 while(SCANF ("%d%d%d", n, m, l) = =3)104 { theEdge_cnt=0;106Memset (Edge,0,sizeof(Edge));107 for(intI=0; i) vect[i].clear ();108 for(intI=0; i)109 { thescanf"%d%d%d",a,b,c);111 if(a==b)Continue; theAdd_node (A,b,c,1);113Add_node (B,a,c,1); the } the cal (); the }117 return 0;118}AC Code UVA 4080

the question.Solution #include #include#include#include#defineMAXN 100005using namespacestd;intN,X,A[MAXN],B[MAXN],C[MAXN];intLowbit (intx) {returnx-x;}voidAddintPosintx) { while(posN) c[pos]=max (c[pos],x), pos+=Lowbit (POS);}intQueryintPOS) { intres=0; while(pos>0) Res=max (Res,c[pos]), pos-=Lowbit (POS); returnRes;} intMain () {scanf ("%d",N); for(intI=1; i) {scanf ("%d", x); a[x]=i;} for(intI=1; i) {scanf ("%d", x); b[x]=i;} for(intI=1; i) {x=query (A[i]); if(X>b[i

The main topic: Give a tree has a root tree, n Group asked, each group asked some key points in the tree, asked to cut off some of the edges so that the root and these points do not connect the minimum cost is how much. The point of total inquiry does not exceed O (n).Idea: The basic idea is that every time you ask to do O (n) DP, which is already fast enough, but there are a lot of inquiries, this is n^2. Note that the points of all inquiries do not add up to O (n), which means that the point o

During this time, I have been learning about andengine and developed a tower guard game "android defensive warfare". Can be learned and used :) The following is an introduction to the game. If you are interested, you can come and play it. Download: The Android world is suffering from attacks from enemies of different races. You need to use the power of the android magic tower to set up defense facilities to defend against enemy intrusion.

Title Link: http://poj.org/problem?id=2892Test instructions: A line of length n, the following m operationD x means the unit x is destroyedR means fix the last destroyed unit.Q X asks the unit and how many contiguous units are around it, and if it's already destroyed, it's 0.Ideas:This problem is the classic line tree entry topic, because only a single point of update, does not involve the interval update, with a tree-like array is more concise.Maintain two tree arrays, one to maintain all 1, an

"Hacker" (Win32.PSWTroj. OnlineGames.73728) I. Threat Level of "legendary hacker" (Win32.Troj. Delf.90112:★ Copy the virus to the system folder, release the virus file, and delete it in batches. A virus is triggered by generating system services. It uses system vulnerabilities to open the user's computer backdoor, control the user's computer, and execute malicious behaviors, such as collecting Computer Information and sending attack commands to attack the specified computer. In addition, the t

, the current growth rate of domestic smartphones has slowed. In recent days, the IDC monthly tracking report shows that July 2014 China's smartphone shipments fell 14.1% yoy. GfK, the research institute, predicts that retail sales in China's smartphone market will reach 393 million in 2014, with a growth rate of just 12% per cent, compared with 82% last year. More and more handset makers are pouring into the market, but the whole handset market is getting smaller, which means there must be a ve

performance of the edge extraction function. When the block_size is set to a larger value, such as block_size=21, 51, etc., it is two value The following is the extraction edgeImport cv2fn= "test3.jpg" Myimg=cv2.imread (FN) Img=cv2.cvtcolor (myimg,cv2. Color_bgr2gray) Newimg=cv2.adaptivethreshold (img,255,cv2. Adaptive_thresh_mean_c,cv2. thresh_binary,5,2) cv2.imshow (' Preview ', newimg) Cv2.waitkey () cv2.destroyallwindows () watermark/2/text/ahr0cdovl2jsb2cuy3nkbi5uzxqvbxloyxnwba==/fon

; - return; A } + intMid= (E[RT].L+E[RT].R) >>1; the if(cmid) { -Update (rt1, c,x); $}Else theUpdate (rt1|1, c,x); thee[rt].ls=e[rt1].ls; thee[rt].rs=e[rt1|1].rs; theE[rt].ms=max (Max (e[rt1].ms,e[rt1|1].ms),e[rt1].rs+e[rt1|1].ls); - if(e[rt1].ls==e[rt1].r-e[rt1].l+1) { ine[rt].ls+=e[rt1|1].ls; the } the if(e[rt1|1].rs==e[rt1|1].r-e[rt1|1].l+1) { Aboute[rt].rs+=e[rt1].rs; the } the } the + intQueryintRtintc) { - if(e[rt].l==e[rt].r| | e[rt].ms==0|| e[rt].

For illustration. Calculates the shortest random sub-ans1. and delete edges. Free to be able to get maximum and short-circuiting between the maximum points ans2 if these two do not communicate. As the shortest distance between two points L.The first idea is to enumerate each edge and then run n the shortest time. However, this complexity is 1000*1000*100*log (100), too large: In fact, at a fixed starting point, the shortest time to find the unit. At the same time can find the shortest single sou

(inti =0; I ) {ret+=D[i]; } returnret;}voidinit () {memset (head,-1,sizeof(head)); ECNT=0; memset (Delta,0,sizeof(delta));}intMain () {//freopen ("In.txt", "R", stdin); intn,m,l; while(~SCANF ("%d%d%d",n,m,m)) {init (); while(m--){ intU,v,w; scanf"%d%d%d",u,v,W); Addedge (--u,--v,w); Addedge (V,U,W); } ll ans1=0; Vectorint>edges; for(intU =0; U ) {Dijkstra (n,u,l,true,-1,-1); ll T=cal (n); Ans1+=T; for(inti =0; I ){ if(~Pa[i]) {Dijkstra (n,u,l,f

 Test instructions: Given an n-node M-edge of the graph, the definition of C for each vertex of the shortest sum, requires the deletion of an edge to re-find a C-value C ', to find the C ' maximum value.Idea: If you use the Floyd algorithm to calculate C, every attempt to delete an edge to be recalculated, the time complexity of O (n*n*n*m), difficult to bear. If Use n times Dijkstra calculates single source shortest circuit, time complexity taste O (N*M*M*LOGN). It looks better than before, b

The so-called irrigation method? is actually a BFS.Note that this question is required to determine whether it is legal, only need to record the number of points passed, to determine whether it is a rectangular area.#include #include #include #include #include using namespace Std;BOOL vis[1005][1005];int n,m,map[1005][1005],flag=0,cnt=0;int dx[]={0,0,1,0,-1},dy[]={0,1,0,-1,0};struct code{int x, y;};Queue Char s[1005];BOOL Judge (int x,int y){if ((x>=1) (xreturn true;return false;}void BFs (int

); Build (M+1R2*rt+1); Pushup (Rt,r-l+1);}voidUpdateintInfointNodeintLintRintRT) { if(l==rnode==l) {segtree[rt].lsum=segtree[rt].rsum=segtree[rt].msum=info; return; } intM= (L+R)/2; if(node2*RT); ElseUpdate (info,node,m+1R2*rt+1); Pushup (Rt,r-l+1);}voidQuary1 (intLintRintLintRintRT) { if(lR) {ans1=ans1+segtree[rt].rsum; if(segtree[rt].rsum!=r-l+1) fail=1; return; } intM= (L+R)/2; if(r>m) Quary1 (l,r,m+1R2*rt+1); if(fail)return; if(l2*RT); if(fail)return;}voidQuary2 (intLintRintLintR

Use find_k_th of the tree array to find the minimum position where sum is K. The time complexity is O (log (n )) /*************************************** **************************************** # Author: neo Fung # Email: neosfung@gmail.com # Last modified: # filename: poj2892 HDU 1540 tunnel warfare. CPP # Description: Use find_k_th of the tree array to find the minimum position where sum is K. The time complexity is O (log (n )) ***************