"bzoj-1176&2683" mokia& simple problem CDQ divided treatment

1176: [Balkan2007]mokia time limit:30 Sec Memory limit:162 MB
submit:1854 solved:821
[Submit] [Status] [Discuss] Description

Maintains a w*w matrix with an initial value of S. Each operation can increase the weight of a lattice, or ask for the total weight of a sub-matrix. Modify the operand m<=160000, asking the number q<=10000,w<=2000000.

Input

The first row of two integers, s,w; S is the matrix initial value; W is the matrix size

Next, each behavior is one of three inputs (without quotation marks):

"1 x y a"

"2 x1 y1 x2 y2"

"3"

Input 1: You need to increase the lattice weights (x, y)

Input 2: You need to ask for the upper left corner (X1,Y1), the lower right corner is (x2,y2) the weight of all the lattice in the matrix, and output

Input 3: Indicates end of input

Output

For each input 2, the output line, that is, the answer to enter 2

Sample Input0 4
1 2 3 3
2 1 1) 3 3
1 2 2 2
2 2 2) 3 4
3Sample Output3
5HINT

Ensure that the answer does not exceed the int range

Source 2683: Simple Question Time limit:50 Sec Memory limit:128 MB
submit:821 solved:339
[Submit] [Status] [Discuss] Description you have a n*n board with an integer in each grid, all at 0 at the beginning, and now you need to maintain two operations:

Command	Parameter limits	Content
1 x y A	1<=x,y<=n,a is a positive integer	Add the numbers in the lattice x, Y, plus a
2 x1 y1 x2 y2	1<=x1<= x2<=n 1<=y1<= y2<=n	Output x1 y1 x2 y2 the number within the rectangle and
3	No	Terminating programs

Input file The first line of a positive integer n. Next, one action per line. Output for each 2 operation, outputs a corresponding answer. Sample Input4
1 2 3 3
2 1 1) 3 3
1 2 2 2
2 2 2) 3 4
3
Sample OUTPUT3
5
hint1<=n<=500000, no more than 200,000 operations, memory limit of 20M. For 100% of the data, the A in operation 1 is not more than 2000. Source

Solution

Can not be directly hard to do, so the use of CDQ division, the general process is:

First, a query operation, split into 4. ++--;

Then all of our operations are sorted, sorted by X-dimension;

Then we set up a tree-like array of y-dimensions, and here we can find that if you press X from small to large, a query is equivalent to the Y-dimension prefix and;

Then CDQ (l,r), enumerate these operations, if you modify the operation sequence number <=mid, then modify, if the query operation ordinal >mid then calculate the answer. This sort and input order by x is not contradictory, you can get the correct answer

And then you have to restore it and then recursively divide it.

Code

BZOJ2683:

#include <iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>using namespaceStd;inlineintRead () {intx=0;CharCh=GetChar ();  while(ch<'0'|| Ch>'9') {ch=GetChar ();}  while(ch>='0'&& ch<='9') {x=x*Ten+ch-'0'; Ch=GetChar ();} returnx;}#defineMAXQ 200010#defineMAXN 500010intS,W,ANS[MAXQ];namespacebit{intTREE[MAXN]; InlineintLowbit (intx) {returnx&-x;} InlinevoidChange (intPosintD) { for(intI=pos; i<=w; I+=lowbit (i)) tree[i]+=D;} InlineintQuery (intPOS) {intRe=0; for(intI=pos; I I-=lowbit (i)) re+=tree[i];returnre;}}using namespaceBIT;structasknode{intId,x,y,opt,del,id; BOOL operator< (ConstAsknode & A)Const        {            return(x==a.x && y==a.y)? Opt<a.opt: ((x==a.x)? y<a.y:x<a.x); }}Q[MAXQ<<2],tmp[maxq<<2];voidCDQ (intLintR) {    if(L==R)return; intMid= (l+r) >>1, z1=l,z2=mid+1;  for(intI=l; i<=r; i++)        {             if(q[i].opt==1&& q[i].id<=mid) Bit::change (Q[i].y,q[i].del); if(q[i].opt==2&& q[i].id>mid) ans[q[i].id]+=bit::query (Q[I].Y) *Q[i].del; }     for(intI=l; i<=r; i++)if(q[i].opt==1&& q[i].id<=mid) Bit::change (q[i].y,-Q[i].del);  for(intI=l; i<=r; i++)if(Q[i].id<=mid) tmp[z1++]=q[i];Elsetmp[z2++]=Q[i];  for(intI=l; i<=r; i++) q[i]=Tmp[i]; CDQ (L,mid); CDQ (Mid+1, R);}intz,t;voidPrework (intX1,intY1,intX2,inty2) {Z++; T++; q[t].id=t; Q[t].id=z; Q[t].x=x1-1; Q[t].y=y1-1; Q[t].del=1; q[t].opt=2; T++; q[t].id=t; Q[t].id=z; q[t].x=x2; Q[t].y=y2; Q[t].del=1; q[t].opt=2; T++; q[t].id=t; Q[t].id=z; Q[t].x=x1-1; Q[t].y=y2; q[t].del=-1; q[t].opt=2; T++; q[t].id=t; Q[t].id=z; q[t].x=x2; Q[t].y=y1-1; q[t].del=-1; q[t].opt=2;}intMain () {W=read (); T=0; z=0;  while(1)        {            intOpt=read ();if(opt==3) Break; if(opt==1) {t++; Q[t].x=read (), Q[t].y=read (), Q[t].del=read (), q[t].id=t,q[t].opt=1;} if(opt==2) {intX1=read (), Y1=read (), X2=read (), y2=read ();        Prework (x1,y1,x2,y2);} } sort (Q+1, q+t+1);//for (int i=1; i<=t; i++) printf ("%d%d%d%d%d\n", q[i].id,q[i].opt,q[i].del,q[i].x,q[i].y);CDQ (1, T);  for(intI=1; i<=z; i++) printf ("%d\n", Ans[i]); return 0;} 

"bzoj-1176&2683" mokia& simple problem CDQ divided treatment