POJ 1195 Mobile phones(二維樹狀陣列)

畫船聽雨發表於2014-02-26
陣列

題意很清晰就是對一個矩陣進行區間的更新與查詢，程式碼寫起來很簡單。就是樹狀陣列存貯方式的原理得知道，感覺樹狀陣列，很強大啊、芳姐給指點的思路，每一個陣列元素的子樹是這個下標二進位制的從右向左第一個1表示之前的元素。比如8->1000所以它可以由0100(4)，0110(6),0111(7)，組成。詳解看之前的部落格：http://blog.csdn.net/fulongxu/article/details/19701281

裸的二維樹狀陣列，注意輸出是：ans = sum (x2+1,y2+1) - sum(x2+1,y1) - sum(x1,y2+1) + sum (x1,y1);減去多於的然後加上減過兩次的。


Mobile phones
Time Limit: 5000MS   Memory Limit: 65536K
Total Submissions: 13686   Accepted: 6348

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 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
#include <algorithm>
#include <iostream>
#include <stdlib.h>
#include <string.h>
#include <iomanip>
#include <stdio.h>
#include <string>
#include <queue>
#include <cmath>
#include <stack>
#include <map>
#include <set>
#define eps 1e-7
#define M 10001000
#define LL __int64
//#define LL long long
#define INF 0x3f3f3f3f
#define PI 3.1415926535898
const int maxn = 1500;

using namespace std;

int c[maxn][maxn];
int n;

int lowbit(int x)
{
    return x&(-x);
}

void add(int x, int y, int ad)
{
    for(int i = x; i <= n; i+= lowbit(i))
        for(int j = y; j<= n; j += lowbit(j))
            c[i][j] += ad;
}

int sum(int x, int y)
{
    int cnt = 0;
    for(int i = x; i > 0; i -= lowbit(i))
        for(int j = y; j > 0; j -= lowbit(j))
            cnt += c[i][j];
    return cnt;
}

int main()
{
    int num;
    while(~scanf("%d",&num))
    {
        if(num == 0)
        {
            scanf("%d",&n);
            memset(c, 0 , sizeof(c));
        }
        else if(num == 1)
        {
            int x, y, ad;
            scanf("%d %d %d",&x, &y, &ad);
            add(x+1, y+1, ad);
        }
        else if(num == 2)
        {
            int l, b, r, t;
            scanf("%d %d %d %d",&l, &b, &r, &t);
            l++; b++; r++; t++;
            int ans = sum(r, t)-sum(l-1, t)-sum(r, b-1)+sum(l-1,b-1);
            printf("%d\n",ans);
        }
        else
            break;
    }
    return 0;
}


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