Codeforces 390 E. Inna and Large Sweet Matrix

主要是樹狀數組的改段求段操作。。。

E. Inna and Large Sweet Matrix time limit per test 1 second memory limit per test 512 megabytes input standard input output standard output

Inna loves sweets very much. That's why she wants to play the "Sweet Matrix" game with Dima and Sereja. But Sereja is a large person, so the game proved small for him. Sereja suggested playing the "Large Sweet Matrix" game.

The "Large Sweet Matrix" playing field is an n?×?m matrix. Let's number the rows of the matrix from 1 to n, and the columns — from 1 tom. Let's denote the cell in the i-th row and j-th column as (i,?j). Each cell of the matrix can contain multiple candies, initially all cells are empty. The game goes in w moves, during each move one of the two following events occurs:

1. Sereja chooses five integers x1,?y1,?x2,?y2,?v (x1?≤?x2,?y1?≤?y2) and adds v candies to each matrix cell (i,?j) (x1?≤?i?≤?x2; y1?≤?j?≤?y2).
2. Sereja chooses four integers x1,?y1,?x2,?y2 (x1?≤?x2,?y1?≤?y2). Then he asks Dima to calculate the total number of candies in cells(i,?j) (x1?≤?i?≤?x2; y1?≤?j?≤?y2) and he asks Inna to calculate the total number of candies in the cells of matrix (p,?q), which meet the following logical criteria: (p?x1 OR p?>?x2) AND (q?y1 OR q?>?y2). Finally, Sereja asks to write down the difference between the number Dima has calculated and the number Inna has calculated (D - I).

   Unfortunately, Sereja's matrix is really huge. That's why Inna and Dima aren't coping with the calculating. Help them!

   Input

   The first line of the input contains three integers n, m and w (3?≤?n,?m?≤?4·106; 1?≤?w?≤?105).

   The next w lines describe the moves that were made in the game.

   • A line that describes an event of the first type contains 6 integers: 0, x1, y1, x2, y2 and v (1?≤?x1?≤?x2?≤?n; 1?≤?y1?≤?y2?≤?m; 1?≤?v?≤?109).
   • A line that describes an event of the second type contains 5 integers: 1, x1, y1, x2, y2 (2?≤?x1?≤?x2?≤?n?-?1; 2?≤?y1?≤?y2?≤?m?-?1).

     It is guaranteed that the second type move occurs at least once. It is guaranteed that a single operation will not add more than 109candies.

     Be careful, the constraints are very large, so please use optimal data structures. Max-tests will be in pretests.

     Output

     For each second type move print a single integer on a single line — the difference between Dima and Inna's numbers.

     Sample test(s) input

     4 5 5
     0 1 1 2 3 2
     0 2 2 3 3 3
     0 1 5 4 5 1
     1 2 3 3 4
     1 3 4 3 4
     

     output

     2
     -21
     

     Note

     Note to the sample. After the first query the matrix looks as:

     22200
     22200 
     00000
     00000
     

     After the second one it is:

     22200
     25500
     03300 
     00000
     

     After the third one it is:

     22201
     25501
     03301
     00001
     

     For the fourth query, Dima's sum equals 5 + 0 + 3 + 0 = 8 and Inna's sum equals 4 + 1 + 0 + 1 = 6. The answer to the query equals 8 - 6 = 2. For the fifth query, Dima's sum equals 0 and Inna's sum equals 18 + 2 + 0 + 1 = 21. The answer to the query is 0 - 21 = -21.

     
     
     

     
     

     
     

     #include 
     #include 
     #include 
     #include 
     
     using namespace std;
     
     typedef long long int LL;
     
     const int maxn=4000400;
     
     int n,m,w,nm;
     LL B[2][maxn],C[2][maxn];
     
     int lowbit(int x)
     {
     	return x&(-x);
     }
     
     void add_b(int id,int p,LL v)
     {
     	for(int i=p;i;i-=lowbit(i)) B[id][i]+=v;
     }
     
     void add_c(int id,int p,LL v)
     {
     	for(int i=p;i<=nm;i+=lowbit(i)) C[id][i]+=v*p;
     }
     
     LL sum_b(int id,int p)
     {
     	LL sum=0;
     	for(int i=p;i<=nm;i+=lowbit(i)) sum+=B[id][i];
     	return sum;
     }
     
     LL sum_c(int id,int p)
     {
     	LL sum=0;
     	for(int i=p;i;i-=lowbit(i)) sum+=C[id][i];
     	return sum;
     }
     
     void ADD(int id,int l,int r,LL v)
     {
     	add_b(id,r,v);
     	add_c(id,r,v);
     	if(l-1)
     	{
     		add_b(id,l-1,-v);
     		add_c(id,l-1,-v);
     	}
     }
     
     LL sUm(int id,int p)
     {
     	if(p)	return sum_b(id,p)*p+sum_c(id,p-1);
     	else return 0;
     }
     
     LL SUM(int id,int l,int r)
     {
     	return sUm(id,r)-sUm(id,l-1);
     }
     
     int main()
     {
     	scanf("%d%d%d",&n,&m,&w);
     	nm=max(n,m)+10;
     	while(w--)
     	{
     		int c,x1,x2,y1,y2,v,dx,dy;
     		scanf("%d%d%d%d%d",&c,&x1,&y1,&x2,&y2);
     		dx=x2-x1+1;dy=y2-y1+1;
     		if(c==0)
     		{
     			scanf("%d",&v);
     			ADD(0,x1,x2,v*dy);
     			ADD(1,y1,y2,v*dx);
     		}
     		else
     		{
     			printf("%I64d\n",SUM(1,y1,y2)-SUM(0,1,x1-1)-SUM(0,x2+1,nm));
     		}
     	}
     	return 0;
     }