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poj1195 Mobile phones

編輯：關於C++

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

Sample Output

這題可以用二維樹狀數組做，感覺二維數組就是一維數組的平方形式，加一個維數，然後循環變為兩個，理解了的話並不難。這裡要注意，矩形的坐標都要加1（即所有坐標都向右上移動一位），這麼做事為了使後面算矩形面積用到getsum(f,g)+getsum(d-1,e-1)-getsum(f,e-1)-getsum(d-1,g)這個公式的時候避免出現d-1=0或者e-1=0，因為lowbit(0)=0,這樣會陷入無限循環，還有就是maxn的大小不能開成2^k，如2048，因為這樣做當i達到2048的時候，數組a[2048][2048]會越界，當然也可以處理一下把數組開到略大於maxn。

#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;
#define maxn 2000
int b[maxn][maxn];
int lowbit(int x){
	return x&(-x);
}

void update(int x,int y,int num)
{
	int i,j;
	for(i=x;i<=maxn;i+=lowbit(i)){
		for(j=y;j<=maxn;j+=lowbit(j)){
			b[i][j]+=num;
		}
	}
}

int getsum(int x,int y)
{
	int num=0,i,j;
	for(i=x;i>0;i-=lowbit(i)){
		for(j=y;j>0;j-=lowbit(j)){
			num+=b[i][j];
		}
	}
	return num;
}

int main()
{
	int n,m,i,j,c,d,f,g,e;
	while(scanf("%d%d",&m,&n)!=EOF)
	{
		memset(b,0,sizeof(b));
		while(1)
		{
			scanf("%d",&c);
			if(c==3)break;
			if(c==1){
			   scanf("%d%d%d",&d,&e,&f);
			   d++;e++;//f++;
			   update(d,e,f);
		    }
		    else if(c==2){
			   scanf("%d%d%d%d",&d,&e,&f,&g);
			   d++;e++;f++;g++;
			   printf("%d\n",getsum(f,g)+getsum(d-1,e-1)-getsum(f,e-1)-getsum(d-1,g));
		    }
		}
		
	}
}