game programming gems 6

;intA[n];intVis[n];intMatch[n];intMpt[n][n];inttot;BOOLisprime[m+Ten];intprime[m+Ten];voidinit () {tot=0; memset (IsPrime,true,sizeof(IsPrime)); isprime[0]=isprime[1]=false; for(intI=2; i) { if(Isprime[i]) prime[tot++]=i; for(intj=0; j) { if(LL) i*prime[j]>m) Break; Isprime[i*prime[j]]=false; if(i%prime[j]==0) Break; } }}intDfsintu) { for(intj=1; j) { if(!vis[j] Mpt[u][j]) {Vis[j]=1; if(match[j]==-1||DFS (Match[j])) {Match[j]=u; retu

}107 }108 109 if(temp==-1)//to add new the {111 node tmp; theTmp.name=TMP2;113Tmp.qmd=Tmp3; the Lis[tmp1].push_back (TMP); theTmp.name=Tmp1; the Lis[tmp2].push_back (TMP);117 if(sex[tmp1]!=SEX[TMP2])118qmd+=Tmp3;119 } - Else if(sex[tmp1]!=SEX[TMP2])121qmd+= (tmp3-temp);122 }123 Else //Ask124coutEndl;; the }126 }127 - return 0;129}not

(Note: "D3d11 game Programming" study Note series by CSDN author Bonchoix wrote, reproduced please indicate the source: Http://blog.csdn.net/BonChoix, thank you ~) Template buffers (stencil buffer) are an off-screen buffer (off-screen buffer) in the same size as the back buffer, which is used primarily to implement some effects. Each pixel in the template buffer is pi,j, and the pixel pi,j in the back buff

#includeusing namespaceStd;typedefstructcannon{intx, y; intR;} Cannon;typedefstructtarget{intx, y;} Target;DoubleCalculate_dis (Cannon m, Target N) {Doubledist = sqrt ((m.x-n.x) * (m.x-n.x) + (M.Y-N.Y) * (M.Y-N.Y)); returnDist;}intMain () {Cannon A, B, C; Target D; intI=0; intR; intCnt=0; intdata[9]; while(scanf_s ("%d", data[i])! = eof i8) {i++; } R= data[0]; A.R=R; A.x= data[1]; A.Y= data[2]; B.R=R; b.x= data[3]; B.y= data[4]; C.R=R; c.x= data[5]; C.y= data[

completely included in the [0,j] between, take the construction of it and do not build it proceeds of the larger. i.e. d (j) =max{d (J), D (j) +X[I].W};(2) When not satisfied (1) But when the J≥X[I].L, we are concerned about the maximum value of X[I].R, after drawing it is easy to know that the value added is to build the income of the pyramid I minus the number of construction area, that is, S (X[I].L,X[I].R)-S (x[i].l,j). The following state transfer equations are obtained: D (X[I].R) =max{d

, representing the number within the collection.Output For each set of data output line, the shape is "case #X: Y". X is the data number, starting with 1, Y is the size of the largest subset.Data range 1≤t≤20 The number 22 in the set S is different and ranges from 1 to 500000. Small Data 1≤n≤15 Big Data 1≤n≤1000 Sample input 352 4 8 16 3252 3 4 6 931 2 3