Line Tree FB

HDU 1166 Enemy Soldiers

The simplest single-point update, because it is the first line of the tree, learning the problem of the Internet after writing

1#include <cstdio>2#include <iostream>3 using namespacestd;4 5 Const intMAXN =50000;6 7 structsegment{8      intl,r,sum;9} t[maxn*4];Ten  One voidBuildintLintRintk) A { -T[K].L =l; -T[K].R =R; theT[k].sum =0; -     if(L = = r)return; -     intMid = (L + r) >>1; -Build (L, Mid,2*k); +Build (mid+1R2*k+1); - } +  A //Enquiry at intans; - voidQueryintKintLintR) - { -     if(T[K].L = = L && T[K].R = =R) -     { -Ans + =t[k].sum; in         return ; -     } to     intMid = (t[k].l + T[K].R) >>1; +     if(r<=mid) Query (2*k,l,r); -     Else if(l>mid) Query (2*k+1, l,r); the     Else *     { $Query2*k,l,mid);Panax NotoginsengQuery2*k+1, mid+1, R); -     } the     return ; + } A  the voidUpdateintKintDintN) + { -     if(T[K].L = = T[K].R && T[K].L = =d) $     { $T[k].sum + = n;//leaf node, updating sum -         return ; -     } the     intMid = (t[k].l + T[K].R) >>1; -     if(d <= mid) update (k*2, d, N);ElseUpdate (k*2+1, D, N);Wuyi        //Update sum of this node theT[k].sum = t[k*2].sum + t[k*2+1].sum; - } Wu  -  About  $ intMain () - { -     intT, Case =0; -scanf"%d",&t); A      while(t--) +     { the         intn,x; -scanf"%d",&n); $Build1N1); the          for(inti =1; I <= N; i++) the         { thescanf"%d",&x); theUpdate1, i,x); -         } inprintf"Case %d:\n",++Case ); the         Charstr[ One]; the          while(~SCANF ("%s", str) && str[0] !='E') About         { the             intb; thescanf"%d%d",&a,&b); the             if(str[0] =='Q') +             { -Ans =0; theQuery1, A, b);Bayiprintf"%d\n", ans); the             } the             if(str[0] =='A') -Update1, A, b); -             if(str[0] =='S') theUpdate1, a,-b); the         } the     } the     return 0; -}

Line Tree FB