【POJ 1195】 Mobile phones (树状数组)

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【POJ 1195】 Mobile phones (树状数组)

Description

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

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

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;
}

【POJ 1195】 Mobile phones (树状数组)

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原文地址：http://blog.csdn.net/challengerrumble/article/details/48314297