POJ 1195 Mobile Phones (two-dimensional tree-like array)

Title Link: POJ 1195 Mobile Phones

"The main topic"

, the start operation is 0 initialized s * S size map with a value of 0

Action 1, enter X Y A, change the value of coordinates (x, y) in the map to a

Operation 2, input L B R T query interval (x, y) l<=x<=r, b<=y<=t, output the rectangle interval and;

Action 3 Closing the program

A typical two-dimensional tree-like array

Two-dimensional tree arrays and one-dimensional tree arrays are the same principle.

We first look at the one-dimensional tree array C "MAXN",

Its storage structure

C1 = A1
C2 = A1 + A2
C3 = A3
C4 = A1 + A2 + A3 + A4
C5 = A5
C6 = A5 + A6


C7 = A7
C8 = A1 + A2 + A3 + A4 + A5 + A6 + A7 + A8
......
C16 = A1 + A2 + A3 + A4 + A5 + A6 + A7 + A8 + A9 + A10 + A11 + A12 + A13 + A14 + A15 + A16
......

The second dimension of the two-dimensional tree array is the same as the normal one-dimensional tree array.

Set the two-dimensional tree array to c[][]

Example: Take a look at the composition of c[][].

Set the original two-dimensional array as:
A[][]={{a11,a12,a13,a14,a15,a16,a17,a18,a19},
{A21,a22,a23,a24,a25,a26,a27,a28,a29},
{A31,a32,a33,a34,a35,a36,a37,a38,a39},
{a41,a42,a43,a44,a45,a46,a47,a48,a49}};
So it corresponds to a two-dimensional tree-like array c[][]?

B[1]={a11,a11+a12,a13,a11+a12+a13+a14,a15,a15+a16,...} This is the first row of a one-dimensional tree array
B[2]={a21,a21+a22,a23,a21+a22+a23+a24,a25,a25+a26,...} This is a one-dimensional tree array of the second row
B[3]={a31,a31+a32,a33,a31+a32+a33+a34,a35,a35+a36,...} This is a one-dimensional, tree-like array of the third row
B[4]={a41,a41+a42,a43,a41+a42+a43+a44,a45,a45+a46,...} This is a one-dimensional tree array of line fourth

So:
C[1][1]=a11,c[1][2]=a11+a12,c[1][3]=a13,c[1][4]=a11+a12+a13+a14,c[1][5]=a15,c[1][6]=a15+a16,...
This is a one-dimensional tree array of the first row of a[][]

c[2][1]=a11+a21,c[2][2]=a11+a12+a21+a22,c[2][3]=a13+a23,c[2][4]=a11+a12+a13+a14+a21+a22+a23+a24,c[2][5]=a15+ A25,c[2][6]=a15+a16+a25+a26,...
This is the tree array after the first row and the second row of the a[][] array are added

C[3][1]=a31,c[3][2]=a31+a32,c[3][3]=a33,c[3][4]=a31+a32+a33+a34,c[3][5]=a35,c[3][6]=a35+a36,...
This is a one-dimensional tree array of the third row of a[][]

C[4][1]=a11+a21+a31+a41,c[4][2]=a11+a12+a21+a22+a31+a32+a41+a42,c[4][3]=a13+a23+a33+a43,...
This is the a[][] array of the first row + the second row + the third row + the fourth row after the tree-like array

attention to the bold part , you can see that the relationship between the storage value between the second dimension and the first dimension between the storage of the relationship is the same,

The number of items stored is lowbit (i).

So modify and query the function, nested two for loop can be completed

void Modify (int a,int b,int val) {for (Int. i=a;i<=n;i+=lowbit (i)) {for (int j=b;j<=n;j+=lowbit (j)) S[i][j]+=val;}} int sum (int x,int y) {int ans=0;for (int i=x;i>0;i-=lowbit (i)) {for (int j=y;j>0;j-=lowbit (j)) Ans+=s[i][j];} return ans;}

The above data source: A tree-like array speaks very good article

"Source Code"

 #include <iostream> #include <string> #include <cstring> #include < cstdio>using namespace Std;const int maxn = 1111;int s[maxn][maxn];int n;int lowbit (int x) {//x binary represents the last 1 position, and of course returns the The value is the binary number consisting of the last 1 and its subsequent 0 return x& (-X);} void Modify (int a,int b,int val) {for (Int. i=a;i<=n;i+=lowbit (i)) {for (int j=b;j<=n;j+=lowbit (j)) S[i][j]+=val;}} int sum (int x,int y) {int ans=0;for (int i=x;i>0;i-=lowbit (i)) {for (int j=y;j>0;j-=lowbit (j)) Ans+=s[i][j];} return ans;} int getsum (int x1,int y1,int x2,int y2) {return sum (x2,y2)-sum (x1-1,y2)-sum (x2,y1-1) +sum (x1-1,y1-1);//First the maximum interval, minus the excess above, The left side is extra, plus the upper left corner minus two times. }int Main () {int op;while (scanf ("%d%d", &op,&n)!=eof) {memset (s,0,sizeof (s)); int Order;int A,b,c,d;while ( scanf ("%d", &order)!=eof && order! = 3) {if (order==1) {scanf ("%d%d%d", &a,&b,&c); Modify (a+1,b+ 1,C);} else {scanf ("%d%d%d%d", &a,&b,&c,&d); int ans = getsum (a+1,b+1,c+1,d+1);p rintf ("%d\n", ans);}} return 0;} 

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POJ 1195 Mobile Phones (two-dimensional tree-like array)