Time Limit: 2000/1000 MS
(Java/Others) Memory Limit: 131072/65536 K
(Java/Others)
Total Submission(s): 1307 Accepted
Submission(s): 674
1 //203MS 4784K 1280 B G++ 2 /* 3 4 题意和原理都和poj 的2155 差不多,这题多了一维而已。 5 6 */ 7 #include<stdio.h> 8 #include<string.h> 9 #define N 105 10 int c[N][N][N]; 11 int lowbit(int i) 12 { 13 return i&(-i); 14 } 15 int update(int x,int y,int z) 16 { 17 int s=0; 18 for(int i=x;i<N;i+=lowbit(i)) 19 for(int j=y;j<N;j+=lowbit(j)) 20 for(int k=z;k<N;k+=lowbit(k)) 21 s+=c[i][j][k]; 22 return s%2; 23 } 24 void getsum(int x,int y,int z) 25 { 26 for(int i=x;i>0;i-=lowbit(i)) 27 for(int j=y;j>0;j-=lowbit(j)) 28 for(int k=z;k>0;k-=lowbit(k)) 29 c[i][j][k]^=1; 30 } 31 int main(void) 32 { 33 int n,m; 34 int op,x1,x2,y1,y2,z1,z2; 35 while(scanf("%d%d",&n,&m)!=EOF) 36 { 37 memset(c,0,sizeof(c)); 38 while(m--){ 39 scanf("%d",&op); 40 if(op==0){ 41 scanf("%d%d%d",&x1,&y1,&z1); 42 printf("%d\n",update(x1,y1,z1)); 43 }else if(op==1){ 44 scanf("%d%d%d%d%d%d",&x1,&y1,&z1,&x2,&y2,&z2); 45 getsum(x1-1,y1-1,z1-1); 46 getsum(x1-1,y1-1,z2); 47 getsum(x1-1,y2,z1-1); 48 getsum(x2,y1-1,z1-1); 49 getsum(x2,y1-1,z2); 50 getsum(x1-1,y2,z2); 51 getsum(x2,y2,z1-1); 52 getsum(x2,y2,z2); 53 } 54 } 55 } 56 return 0; 57 }
hdu 3584 Cube (三维树状数组),布布扣,bubuko.com
原文:http://www.cnblogs.com/GO-NO-1/p/3639300.html