POJ 1195 Mobile phones（ IOI 2001） 二维树状数组 CDQ分治 代码
Mobile phones
| Time Limit: 5000MS | Memory Limit: 65536K | |
| Total Submissions: 24928 | Accepted: 11377 |
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
二维树状数组代码：
#include
#include
#include
#include
#include
using namespace std;
#define MAXN 1025
int N,S[MAXN][MAXN];
void add(int x,int y,int value)
{
for(int i=x;i<=N;i+=(i&(-i)))
for(int j=y;j<=N;j+=(j&(-j)))
S[i][j]+=value;
}
int getsum(int x,int y)
{
int tmp=0;
for(int i=x;i>=1;i-=(i&(-i)))
for(int j=y;j>=1;j-=(j&(-j)))
tmp+=S[i][j];
return tmp;
}
int main()
{
//freopen("input.in","r",stdin);
//freopen("output.out","w",stdout);
int sign;
while(scanf("%d",&sign)!=EOF)
{
if(sign==0)
{
scanf("%d",&N);
continue;
}
if(sign==1)
{
int X,Y,V;
scanf("%d%d%d",&X,&Y,&V);
X++,Y++;
add(X,Y,V);
}
if(sign==2)
{
int X1,Y1,X2,Y2;
scanf("%d%d%d%d",&X1,&Y1,&X2,&Y2);
X1++,Y1++,X2++,Y2++;
printf("%d\n",getsum(X1-1,Y1-1)-getsum(X1-1,Y2)+getsum(X2,Y2)-getsum(X2,Y1-1));
}
if(sign==3) break;
}
return 0;
} CDQ分治代码：
#include
#include
#include
#include
#include
using namespace std;
#define MAX 640001
#define MAXN 2000001
#define MAXQ 60001
struct ASK{int num,x,y,v,kind,in;}a[MAX],T[MAX];
int N,Q,tot=0,S[MAXN],ans[MAXQ];
void add(int x,int value)
{
for(int i=x;i<=N;i+=(i&(-i)))S[i]+=value;
}
int getsum(int x){int tmp=0;for(int i=x;i>0;i-=(i&(-i)))tmp+=S[i];return tmp;};
bool cmp(ASK X,ASK Y){return X.x CDQ分治的详细分析可见：https://blog.csdn.net/csyzcyj/article/details/107298731
转载注明出处：https://blog.csdn.net/csyzcyj/