BZOJ 1176([Balkan2007]Mokia-CDQ分治-分治询问)

 

1176: [Balkan2007]Mokia

Time Limit: 30 Sec   Memory Limit: 162 MB
Submit: 185   Solved: 94
[ Submit][ Status]

Description

维护一个W*W的矩阵，每次操作可以增加某格子的权值，或询问某子矩阵的总权值。 修改操作数M<=160000，询问数Q<=10000，W<=2000000。

Input

 

Output

 

Sample Input

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

Sample Output

3
5

HINT

Input file Output file Meaning
0 4 Table size is , filled with zeroes.

1 2 3 3 Add 3 customers at (2, 3).
2 1 1 3 3 Query sum of rectangle , .

3 Answer.
1 2 2 2 Add 2 customers at (2, 2).
2 2 2 3 4 Query sum of rectangle , .

5 Answer
3 Exit your program.

 

 

裸CDQ...1A

注意几个关键部分的写法

步骤：

1.读入询问，按x排序

2.将[L,R]中的数分为前部分操作，后部分操作（各部分仍保持X升序）

3.将前面对后面的影响记录ans

4.复原影响

5.递归[L,M],[M+1,R]

 

#include<cstdio>

#include<cstdlib>

#include<cstring>

#include<iostream>

#include<algorithm>

#include<functional>

#include<cmath>

#include<cctype>

#include<cassert>

#include<climits>

using namespace std;

#define For(i,n) for(int i=1;i<=n;i++)

#define Rep(i,n) for(int i=0;i<n;i++)

#define Fork(i,k,n) for(int i=k;i<=n;i++)

#define ForD(i,n) for(int i=n;i;i--)

#define Forp(x) for(int p=pre[x];p;p=next[p])

#define RepD(i,n) for(int i=n;i>=0;i--)

#define MEM(a) memset(a,0,sizeof(a))

#define MEMI(a) memset(a,127,sizeof(a))

#define MEMi(a) memset(a,128,sizeof(a))

#define INF (2139062143)

#define F (1000000009)

#define MAXM (640000+10)

#define MAXQ (10000+10)

#define MAXN (2000000+10) 

typedef long long ll;

int n,m,q;

ll ans[MAXQ];

struct comm

{

   int no,x,y,v,q,type;

   comm():no(0),x(0),y(0),v(0),q(0),type(0){}

   comm(int _no,int _x,int _y,int _v,int _q,int _type):no(_no),x(_x),y(_y),v(_v),q(_q),type(_type){}

   friend bool operator<(comm a,comm b){return a.x<b.x;}      

}ask[MAXM];

struct arr_tree

{

   ll a[MAXN];

   arr_tree(){memset(a,0,sizeof(a));}

   void add(int x,ll c)

   {

      for(int i=x;i<=n;i+=i&(-i)) a[i]+=c;

   }

   int qur(int x)

   {

      ll ans=0;

      for(int i=x;i;i-=i&(-i)) ans+=a[i];

      return ans;   

   }

   void clear()

   {

      memset(a,0,sizeof(a));

   }

}T;

comm tmp[MAXM];

void solve(int l,int r)

{

   if (l==r) return;

   int m=l+r>>1;

   int s1=l-1,s2=m;

   Fork(i,l,r) tmp[ask[i].no<=m?++s1:++s2]= ask[i];

   memcpy(ask+l,tmp+l,sizeof(comm)*(r-l+1));

   int now=l;

   Fork(i,m+1,r) //遍历询问 

   {

      if (ask[i].type==2)

      {

         while (ask[now].x<=ask[i].x&&now<=m)

         {

            if (ask[now].type==1) T.add(ask[now].y,ask[now].v);

            now++;

         }

         ans[ask[i].q]+=ask[i].v*T.qur(ask[i].y);         

      }

   }

   now--;

   while (l<=now) {if (ask[now].type==1) T.add(ask[now].y,-ask[now].v);now--;}

   solve(l,m),solve(m+1,r);

}

bool work()

{

   scanf("%d",&n);

   int type,no=0,x,y,x1,y1,x2,y2,v,q=0;

   memset(ans,0,sizeof(ans));T.clear();

   while (scanf("%d",&type))

   {

      if (type==0||type==3) break;

      else if (type==1)

      {

         scanf("%d%d%d",&x,&y,&v);

         no++;ask[no]=comm(no,x,y,v,0,1);

      }

      else if (type==2)

      {

         scanf("%d%d%d%d",&x1,&y1,&x2,&y2);q++;

         no++;ask[no]=comm(no,x1-1,y1-1,1,q,2);

         no++;ask[no]=comm(no,x2,y2,1,q,2);

         no++;ask[no]=comm(no,x1-1,y2,-1,q,2);

         no++;ask[no]=comm(no,x2,y1-1,-1,q,2);         

      }

   }

   sort(ask+1,ask+1+no);

   solve(1,no);

   For(i,q) cout<<ans[i]<<endl;

   return type==0;

}

int main()

{

   //freopen("bzoj1176.in","r",stdin);

   int type;

   scanf("%d",&type);if (type==3) return 0;

   while (work());

   //cout<<"END"<<endl;while (1);

   return 0;

}