本文主要是介绍树状数组模板+poj1195（二维树状数组），希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
感谢学长的博客~~http://blog.csdn.net/lin375691011/article/details/21247409
在数组长度为n的树状数组中：
寻找下一个需要添加的数的下标：
int lowbit(int x)
{ return x&(-x);
}
void add(int x,int val)
{ for(;x<=n;x+=lowbit(x)) { num[x]+=val; }
}
void add(int x,int y,int val)
{ for(int i=x;i<=s;i+=lowbit(i)) { for(int j=y;j<=s;j+=lowbit(j)) { num[i][j]+=val; } }
} 一维树状数组查询是这样的：
int query(int x)
{ int ans=0; for(;x>0;x-=lowbit(x)) { ans+=c[i]; } return ans;
}
int query(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+=num[i][j]; } } return ans;
} 下面看一个纯二维的树状数组模板题
Mobile phones
| Time Limit: 5000MS | Memory Limit: 65536K | |
| Total Submissions: 18744 | Accepted: 8647 |
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.
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
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 4 1 1 2 3 2 0 0 2 2 1 1 1 2 1 1 2 -1 2 1 1 2 3 3
Sample Output
3 4
Source
题目大意：
给定矩阵大小，可以更新矩阵中的某些数，要求输出某个范围内的所有数之和。
#include<stdio.h>
#include<string.h>
#define lowbit(x) x&(-x)
#define maxn 1025
int c[maxn][maxn];
int n;
void add(int x,int y,int a)
{for( int i = x; i <=n; i += lowbit(i)){for( int j = y; j <=n; j += lowbit(j)){c[i][j]+=a;}}
}
int sum( int l,int r )
{int ans = 0;for( int i = l; i > 0; i -= lowbit(i) ){for( int j = r; j > 0; j -=lowbit(j)){ans+=c[i][j];}}return ans;
}
int main()
{int i,j,k;scanf("%d%d",&k,&n);memset(c,0,sizeof(c));int op;while(~scanf("%d",&op)){if(op>=3) break;if(op == 1){int x,y,a;scanf("%d%d%d",&x,&y,&a);add(x+1,y+1,a);}else if(op == 2){int l,r,b,t;scanf("%d%d%d%d",&l,&b,&r,&t);l++,b++,r++,t++;printf("%d\n",sum(r,t) - sum(r,b-1) - sum(l-1, t) + sum(l-1, b-1));}}
}
这篇关于树状数组模板+poj1195（二维树状数组）的文章就介绍到这儿，希望我们推荐的文章对编程师们有所帮助！
