Mobile phones 【典型的典型树状数组】

 Mobile phones

Time Limit: 5000MS Memory Limit: 65536KTotal Submissions: 5865 Accepted: 2447

Description

Suppose that the fourth generation mobile phone base stations in the Tampere area operate as follows. The area is divided into squares. The squares form an S * S matrix with the rows and columns numbered from 0 to S-1. Each square contains a base station. The number of active mobile phones inside a square can change because a phone is moved from a square to another or a phone is switched on or off. At times, each base station reports the change in the number of active phones to the main base station along with the row and the column of the matrix.

Write a program, which receives these reports and answers queries about the current total number of active mobile phones in any rectangle-shaped area.

Input

The input is read from standard input as integers and the answers to the queries are written to standard output as integers. The input is encoded as follows. Each input comes on a separate line, and consists of one instruction integer and a number of parameter integers according to the following table.

The values will always be in range, so there is no need to check them. In particular, if A is negative, it can be assumed that it will not reduce the square value below zero. The indexing starts at 0, e.g. for a table of size 4 * 4, we have 0 <= X <= 3 and 0 <= Y <= 3.

Table size: 1 * 1 <= S * S <= 1024 * 1024
Cell value V at any time: 0 <= V <= 32767
Update amount: -32768 <= A <= 32767
No of instructions in input: 3 <= U <= 60002
Maximum number of phones in the whole table: M= 2^30

Output

Your program should not answer anything to lines with an instruction other than 2. If the instruction is 2, then your program is expected to answer the query by writing the answer as a single line containing a single integer to standard output.

Sample Input

0 41 1 2 32 0 0 2 2 1 1 1 21 1 2 -12 1 1 2 3 3

Sample Output

34

我写 的其实是一维的，多个一维的搞起来 。时间竟然4000MS ，，，

还没想出来二维的  继续中 先把以为的思路写这里
一行一维的树状数组做
那么就是m个一维树状数组，这里的左边都有限制，所以我就减去左边的

效率很低这个代码：

#include   <stdio.h>   #include   <string.h>   int e,tree[1030][1030];void updata(int x,int y,int A){while(y<=e){tree[x][y]+=A;y+=y&(-y);}}int read(int x,int y){int r=0;while(y>0){//printf("tree[%d][%d] = %d/n",x,y,tree[x][y]);r+=tree[x][y];y-=y&(-y);}return r;}int main(){int s,t,re;int x,y,A,l,r,d,u;scanf("%d%d",&s,&e);memset(tree,0,sizeof(tree));while(1){re=0;scanf("%d",&t);if(t==3)break;if(t==1){scanf("%d%d%d",&x,&y,&A);x++;y++;updata(x,y,A);}if(t==2){scanf("%d%d%d%d",&l,&d,&r,&u);l++;d++;r++;u++;for(int i=l;i<=r;i++){re+=read(i,u);re-=read(i,d-1);}printf("%d/n",re);}}return 0;}

典型的二维树状数组的应用。

以前有几篇论文讨论了如何要用二维线段树来做，相当麻烦。不过二维树状数组则十分简单。

不过树状数组的应用范围很窄，“在一个N*N的区域内，可以在某个点增加或减少A，并动态询问一个矩形区域内的数值之和 “用树状数组更好。
但

如果修改题目为“可以给某个矩形区域每个点增加或减少A，

并动态询问某个点的数值”那么树状数组就不行了，而线段树可以使用。

//////////

修改成二维就是

void updata(int x,int y,int A){int ty = y;while(x<=e){y=ty;while(y<=e){tree[x][y]+=A;y+=y&(-y);}x+=x&(-x);}}int read(int x,int y){int r=0,ty=y;while(x>0){y=ty;while(y>0){//printf("tree[%d][%d] = %d/n",x,y,tree[x][y]);r+=tree[x][y];y-=y&(-y);}x-=x&(-x);}return r;} 

然后读的时候是

保证坐标都从1开始，否则开始都加+1。

result = read(D(x4,x4))+read(A(x1-1,y1-1))-read(B(x2-1,y2))-read(C(x3,y3-1));

看起来视乎像对角相加，再减去另外对角的和。没有证明，姑且先这样记了。

好了看二维做的代码（这题原本就要用二维的）：


 #include   <stdio.h>   #include   <string.h>   int e,tree[1030][1030];void updata(int x,int y,int A){int ty = y;while(x<=e){y=ty;while(y<=e){tree[x][y]+=A;y+=y&(-y);}x+=x&(-x);}}int read(int x,int y){int r=0,ty=y;while(x>0){y=ty;while(y>0){//printf("tree[%d][%d] = %d/n",x,y,tree[x][y]);r+=tree[x][y];y-=y&(-y);}x-=x&(-x);}return r;}int main(){int s,t,re;int x,y,A,l,r,d,u;scanf("%d%d",&s,&e);memset(tree,0,sizeof(tree));while(1){re=0;scanf("%d",&t);if(t==3)break;if(t==1){scanf("%d%d%d",&x,&y,&A);x++;y++;updata(x,y,A);}if(t==2){scanf("%d%d%d%d",&l,&d,&r,&u);l++;d++;r++;u++;re=read(r,u)-read(l-1,u)-read(r,d-1)+read(l-1,d-1);printf("%d/n",re);}}return 0;} 

ccy觉得，我们应该无视掉那个诡异名字“树状数组”，脑子里不要想太多，只需要知道，

它能够通过某种简单而快速的方法，算出某个字串的和就行了。

CCY 是这句话的原作者。