poj1195 (二维树状数组）

http://poj.org/problem?id=1195

Mobile phones

Time Limit: 5000MS Memory Limit: 65536KTotal Submissions: 15187 Accepted: 7016

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 41 1 2 32 0 0 2 2 1 1 1 21 1 2 -12 1 1 2 3 3

Sample Output

34

#include <iostream>#include<cstdio>#include <vector>#include <cstring>using namespace std;#define maxn 1100int C[maxn][maxn];int Lowbit[maxn];int s;void Add( int y, int x,int a){    while( y <= s )    {        int tmpx = x;        while( tmpx <= s )        {            C[y][tmpx] += a;            tmpx += Lowbit[tmpx];        }        y += Lowbit[y];    }}int QuerySum( int y, int x)//查询第1行到第y行，第1列到第x列的和{    int nSum = 0;    while( y > 0 )    {        int tmpx = x;        while( tmpx > 0)        {            nSum += C[y][tmpx];            tmpx -= Lowbit[tmpx];        }        y -= Lowbit[y];    }    return nSum;}int main(){    int cmd,x,y,a,l,b,r,t;    for( int i = 1; i <= maxn; i ++ ) Lowbit[i] = i & (-i);    while(true)    {        scanf("%d",&cmd);        switch( cmd)        {        case 0:            scanf("%d",& s);            memset( C,0,sizeof(C));            break;        case 1:            scanf("%d%d%d",&x ,&y,&a);            Add( y + 1, x + 1, a);            break;        case 2:            scanf("%d%d%d%d",&l , &b, &r,&t);            l ++;            b++;            r ++;            t ++;            printf("%d\n",QuerySum(t,r) +QuerySum(b-1,l-1) - QuerySum(t,l-1) - QuerySum(b-1,r));            break;        case 3:            return 0;        }    }}

#include <iostream>#include<cstdio>#include<cstring>using namespace std;#define maxn 1100int Tree[maxn * 3][maxn * 3];//二维线段树，每一行都是一棵完全二叉树，用于存放一棵x方向线段树，树节点只存放 对应的区间（矩阵）的数字之和int S; //矩阵宽度void Add_x( int rooty,int rootx, int L, int R, int x ,int a)//tree[rooty][rootx]对应的矩阵x方向上范围是[L,R]{    Tree[rooty][rootx] += a;    if( L == R )        return;    int mid = (L + R )/2;    if( x <= mid )        Add_x(rooty,( rootx << 1) + 1, L ,mid, x, a);    else        Add_x(rooty,( rootx << 1) + 2, mid + 1,R, x, a);}void Add_y(int rooty, int L,int R, int y, int x,int a)//tree[rooty][rootx]对应的矩阵y方向上范围是[L,R]{    Add_x( rooty,0, 1, S, x,a);    if( L == R)        return;    int mid = (L + R )/2;    if( y <= mid )        Add_y( ( rooty << 1) + 1, L, mid,y, x, a);    else        Add_y( ( rooty << 1) + 2, mid+1, R, y, x, a);}int QuerySum_x(int rooty, int rootx, int L, int R ,int x1,int x2){    if( L == x1 && R == x2)        return Tree[rooty][rootx];    int mid = ( L + R ) /2 ;    if( x2 <= mid )        return QuerySum_x( rooty, (rootx << 1) + 1,                           L, mid,x1,x2);    else if( x1 > mid )        return QuerySum_x( rooty, (rootx << 1) + 2,                           mid+1,R, x1,x2);    else        return QuerySum_x( rooty,(rootx << 1) + 1,                           L, mid ,x1,mid) +               QuerySum_x( rooty,(rootx << 1) + 2,                           mid + 1, R, mid + 1,x2);}int QuerySum_y(int rooty, int L, int R ,int y1, int y2, int x1,int x2){    if( L == y1 && R == y2 )        return QuerySum_x(rooty,0,1,S,x1,x2);    int mid = ( L + R ) /2;    if( y2 <= mid )        return QuerySum_y((rooty << 1)+1,L,mid,y1,y2,x1,x2);    if( y1 > mid )        return QuerySum_y((rooty<<1)+2,mid +1,R,y1,y2,x1,x2);    else        return QuerySum_y((rooty << 1) + 1,L,mid,y1,mid,x1,x2)+QuerySum_y((rooty <<1)+2,mid+1,R,mid +1,y2,x1,x2);}int main(){    int cmd,x,y,a,l,b,r,t;    while( true)    {        scanf("%d",&cmd);        switch( cmd)        {        case 0:            scanf("%d",&S);            memset( Tree,0,sizeof(Tree));            break;        case 1:            scanf("%d%d%d",&x ,&y,&a);            Add_y(0, 1,S, y + 1, x + 1, a);            break;        case 2:            scanf("%d%d%d%d",&l , &b, &r,&t);            l ++;            b++;            r ++;            t ++;            printf("%d\n",QuerySum_y(0,1,S,b,t,l,r));            break;        case 3:            return 0;        }    }}