POJ 1195 Mobile phones （二维树状数组）

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 

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

3
4

二维树状数组求和与改动，从一维能够扩展到二维。

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <set>
#include <stack>
#include <cctype>
#include <algorithm>
#define lson o<<1, l, m
#define rson o<<1|1, m+1, r
using namespace std;
typedef long long LL;
const int mod = 99999997;
const int MAX = 1000000000;
const int maxn = 100005;
int ca, s, op;
int c[1100][1100];
void add(int i, int j, int v) {
    while(i <= s) {
        int y = j;
        while(y <= s) {
            c[i][y] += v;
            y += y&-y;
        }
        i += i&-i;
    }
}
int query(int i ,int j) {
    int ans = 0;
    while(i > 0) {
        int y = j;
        while(y > 0) {
            ans += c[i][y];
            y -= y&-y;
        }
        i -= i&-i;
    }
    return ans;
}
int main()
{
    cin >> ca >> s;
    while(1) {
        scanf("%d", &op);
        if(op == 1) {
            int a, b, v;
            scanf("%d%d%d", &a, &b, &v);
            add(a+1, b+1, v);
        } else if(op == 2) {
            int x1, x2, y1, y2;
            scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
            int ans = query(x2+1, y2+1) + query(x1, y1) - query(x2+1, y1) - query(x1, y2+1);
            printf("%d\n", ans);
        } else break;
    }
    return 0;
}