【POJ 1195】 Mobile phones (树状数组)

【POJ 1195】 Mobile phones (树状数组)

Mobile phones

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

Source

IOI 2001

二维树状数组 一前光写过一维的 一开始写的是第二维树状数组 第一维暴力。。。跑了四千多。。

后来查了查第一维跟第二维一样 累加Lowbite即可 查询的时候用大矩阵减去上面横向矩阵和左面纵向 这样多减了左上角的小矩阵 再加上即可 速度快了好多

兴起封了个类玩。。

代码如下:

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cstdlib>

using namespace std;

class Trie
{
public:
    Trie()
    {
        memset(tr,0,sizeof(tr));
    }
    void run(int t)
    {
        int a,b,c,d;
        if(!t)
        {
            scanf("%d",&a);
            n = a;
        }
        else if(t == 1)
        {
            scanf("%d %d %d",&a,&b,&c);
            Add(a+1,b+1,c);
        }
        else
        {
            scanf("%d %d %d %d",&a,&b,&c,&d);
            printf("%d\n",Sum(c+1,d+1)+Sum(a,b)-Sum(a,d+1)-Sum(c+1,b));
        }
    }
private:
    int tr[1033][1033],n;
    int Lowbite(int x)
    {
        return x&(-x);
    }
    void Add(int x,int y,int data)
    {
        int i,j;
        for(i = x; i <= n; i += Lowbite(i))
            for(j = y; j <= n; j += Lowbite(j))
                tr[i][j] += data;
    }
    int Sum(int x,int y)
    {
        int ans = 0;
        int i,j;
        for(i = x; i > 0; i -= Lowbite(i))
            for(j = y; j > 0; j -= Lowbite(j))
                ans += tr[i][j];
        return ans;
    }
};

int main()
{
    int t;
    Trie *tr = new Trie();
    while(~scanf("%d",&t) && t != 3)
    {
        tr->run(t);
    }
    return 0;
}

//原
#include <iostream>
#include <cstring>
#include <cstdio>
#include <cstdlib>

using namespace std;

int tr[1033][1033],n;

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

void Add(int x,int y,int data)
{
    int i,j;
    for(i = x; i <= n; i += Lowbite(i))
        for(j = y; j <= n; j += Lowbite(j))
            tr[i][j] += data;
}

int Sum(int x,int y)
{
    int ans = 0;
    int i,j;
    for(i = x; i > 0; i -= Lowbite(i))
        for(j = y; j > 0; j -= Lowbite(j))
            ans += tr[i][j];
    return ans;
}

int main()
{
    int t,a,b,c,d;
    memset(tr,0,sizeof(tr));
    while(~scanf("%d",&t) && t != 3)
    {
        if(!t)
        {
            scanf("%d",&a);
            n = a;
        }
        else if(t == 1)
        {
            scanf("%d %d %d",&a,&b,&c);
            Add(a+1,b+1,c);
        }
        else
        {
            scanf("%d %d %d %d",&a,&b,&c,&d);
            printf("%d\n",Sum(c+1,d+1)+Sum(a,b)-Sum(a,d+1)-Sum(c+1,b));
        }
    }
    return 0;
}

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