POJ 1195 二维树状数组
Time Limit: 5000MS | Memory Limit: 65536K | |
Total Submissions: 18489 | Accepted: 8558 |
Description
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
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
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
//树状数组模板
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int flag,X,Y,A,L,B,R,T,S;
int squ[][];
int lowbit(int x)
{
return x&(-x);
}
void add(int idx,int idy,int val)
{
for(int i=idx;i<=;i+=lowbit(i))
{
for(int j=idy;j<=;j+=lowbit(j))
{
squ[i][j]+=val;
}
}
}
int sum(int idx,int idy)
{
int s=;
for(int i=idx;i>;i-=lowbit(i))
{
for(int j=idy;j>;j-=lowbit(j))
{
s+=squ[i][j];
}
}
return s;
}
int ans(int x1,int y1,int x2,int y2)
{
return (sum(x2,y2)-sum(x1-,y2)-sum(x2,y1-)+sum(x1-,y1-));
}
int main()
{
while(scanf("%d",&flag)&&flag!=)
{
if(flag==)
{
scanf("%d",&S);
memset(squ,,sizeof(squ));
}
else if(flag==)
{
scanf("%d%d%d",&X,&Y,&A);
if(A>=)
add(X+,Y+,A);
else
{
int num=ans(X+,Y+,X+,Y+);
add(X+,Y+,num+A>?A:-num);
}
}
else if(flag==)
{
scanf("%d%d%d%d",&L,&B,&R,&T);
printf("%d\n",ans(L+,B+,R+,T+));
}
}
return ;
}
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