描述
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.
输入
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
输出
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.
样例输入
0 4
1 1 2 3
2 0 0 2 2
1 1 1 2
1 1 2 -1
2 1 1 2 3
3
样例输出
3
4
解析： 这个因为操作次数和操作的空间相乘比较大，所以放弃一般的朴素做法，然后看题目，单点操作，区间询问。考虑树状数组，首先我是把二维化成一维写的一维的树状数组，就是相当于把二维数组看做成一维，ps:当时不会二维，现在会了。
首先是一维的代码，875ms
#include<bits/stdc++.h> using namespace std; typedef long long ll; const int N=1100; int c[N*N],n; inline int lowbit(int n) {return n&(-n);} void update(int x,int y) { for(int i=x;i<=n*n;i+=lowbit(i)){ c[i]+=y;} } int sum(int x){ int ans=0; for(int i=x;i;i-=lowbit(i)) ans+=c[i]; return ans; } int main() { int t,op; cin>>t>>n; while(scanf("%d",&op),op!=3) { if(op==1) { int x,y,a; scanf("%d%d%d",&x,&y,&a); x+=1,y+=1; update((x-1)*n+y,a); } else if(op==2) { int L,R,B,T; scanf("%d%d%d%d",&L,&B,&R,&T); L++,B++,R++,T++;//防止出现lowbit(0)=0,卡死 int ans=0; for(int i=L;i<=R;++i) { ans+=(sum((i-1)*n+T)-sum((i-1)*n+B-1));//容易出现一个问题， } printf("%d\n",ans); } } return(0); }
然后是二维的代码 218ms ，然后我说下二维是怎么理解，你可以把一个一维的数组看成普通一维树状数组的一个元素
#include<cstring> #include<cstdio> #include <iostream> #include<algorithm> using namespace std; const int N=1025; int c[N][N],s; void update(int x,int y,int a) { for(int i=x;i<=s;i+=i&(-i)) for(int j=y;j<=s;j+=j&(-j)) c[i][j]+=a; } int sum(int x,int y) { int res=0; for(int i=x;i;i-=i&(-i)) for(int j=y;j;j-=j&(-j)) res+=c[i][j]; return res; } int main() { int t,op; cin>>t>>s; while(scanf("%d",&op),op!=3) { if(op==1) { int x,y,a; scanf("%d%d%d",&x,&y,&a); x+=1,y+=1; update(x,y,a); } else if(op==2) { int L,R,B,T; scanf("%d%d%d%d",&L,&B,&R,&T); L++,B++,R++,T++;//防止出现lowbit(0)=0,卡死 int ans=0; ans=sum(R,T) + sum(L-1,B-1) - sum(L-1,T) - sum(R,B-1); printf("%d\n",ans); } } return(0); }
7/25 Mobile phones(树状数组/二维树状数组)
原文：https://www.cnblogs.com/NoahBBQ/p/15058814.html