HDU 3584-cube (three-dimensional bit)

Test instructions

Give you three-dimensional space two operations, give two vertex coordinates, they determine the range (box) within the number of all the inverse, query to the value of the point. Initially all zeros

Analysis:

With the knowledge of the previous, using bit to achieve the interval update single-point query, and then use multidimensional implementation can

#include <map>#include<Set>#include<list>#include<cmath>#include<queue>#include<stack>#include<cstdio>#include<vector>#include<string>#include<cctype>#include<complex>#include<cassert>#include<utility>#include<cstring>#include<cstdlib>#include<iostream>#include<algorithm>using namespaceStd;typedef pair<int,int>Pii;typedefLong Longll;#defineLson l,m,rt<<1#definePi ACOs (-1.0)#defineRson m+1,r,rt<<11#defineAll 1,n,1#defineRead Freopen ("In.txt", "R", stdin)#defineN 110Constll infll =0x3f3f3f3f3f3f3f3fll;Const intinf=0x7ffffff;Const intMoD =1000000007;inta[n][n][n],n,m;intLowbit (intx) {    returnx& (-x);}voidAddintXintYintz) {     for(inti=x;i<=n;i+=lowbit (i)) for(intj=y;j<=n;j+=Lowbit (j)) for(intk=z;k<=n;k=k+Lowbit (k)) A[i][j][k]++;//statistical parity +1,-1}intSumintXintYintz) {    intnum=0;  for(intI=x;i>0; i-=lowbit (i)) for(intJ=y;j>0; j-=Lowbit (j)) for(intK=z;k>0; k-=Lowbit (k)) Num+=A[i][j][k]; returnnum;}intMain () {intop,x1,y1,z1,x2,y2,z2;  while(~SCANF ("%d%d",&n,&m)) {memset (A,0,sizeof(a));  while(m--) {scanf ("%d",&op); if(op==1) {scanf ("%d%d%d%d%d%d",&x1,&y1,&z1,&x2,&y2,&z2);            Add (X1,Y1,Z1); Add (x1,y1,z2+1); Add (x2+1, Y1,Z1); Add (X1,y2+1, Z1); Add (x2+1, y2+1, Z1); Add (x2+1, y1,z2+1); Add (X1,y2+1, z2+1); Add (x2+1, y2+1, z2+1); }        Else{scanf ("%d%d%d",&x1,&y1,&Z1); printf ("%d\n", SUM (X1,Y1,Z1)%2); }      }    }return 0;}

HDU 3584-cube (three-dimensional bit)