CDQ求子矩阵的和

Posted 2022-09-22 shuaihui520

Description
维护一个W*W的矩阵，初始值均为S.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
第一行两个整数,S,W;其中S为矩阵初始值;W为矩阵大小

接下来每行为一下三种输入之一(不包含引号):

"1 x y a"

"2 x1 y1 x2 y2"

"3"

输入1:你需要把(x,y)(第x行第y列)的格子权值增加a

输入2:你需要求出以左下角为(x1,y1),右上角为(x2,y2)的矩阵内所有格子的权值和,并输出

输入3:表示输入结束

Output
对于每个输入2,输出一行,即输入2的答案

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
 

保证答案不会超过int范围

 

#include<cstdio>
#include<cmath>
#include<cstring>
#include<string>
#include<cstdlib>
#include<algorithm>
#include<iostream>
#include<queue>
#include<stack>
#include<map>
 
using namespace std;
 
#define FOU(i,x,y) for(int i=x;i<=y;i++)
#define FOD(i,x,y) for(int i=x;i>=y;i--)
#define MEM(a,val) memset(a,val,sizeof(a))
#define PI acos(-1.0)
 
const double EXP = 1e-9;
typedef long long ll;
typedef unsigned long long ull;
const int INF = 0x3f3f3f3f;
const ll MINF = 0x3f3f3f3f3f3f3f3f;
const double DINF = 0xffffffffffff;
const int mod = 1e9+7;
const int N = 2e6+5;
 
struct node
    int typ;     //类型
    int x,y;
    int id;     //记录输入相对位置
    int val;
    int pos;     //输出答案用
a[N],tmp[N];
 
int n,w,tot;
int ans[10005];   //答案数组
 
bool cmp(node a,node b)
    if(a.x==b.x&&a.y==b.y)
        return a.id<b.id;
    if(a.x==b.x)
        return a.y<b.y;
    return a.x<b.x;

 
 
int tree[N]; //tree数组按二进制存,根据n的末尾0的个数存取,树状数组
 
int lowbit(int x)

    return x&(-x);

 
int Query(int x)  //返回1到x的前缀和

    int res=0;
    while(x)
    
        res+=tree[x];
        x-=lowbit(x);
    
    return res;

 
void Add(int x,int v)  //实现a[x]+v;

    while(x<=n)        //注意这里是小于等于k,还是n,k是数据范围
    
        tree[x]+=v;
        x+=lowbit(x);
    

 
void clearr(int x)
    while(x<=n)
        if(tree[x]==0)
            break;
        tree[x]=0;
        x+=lowbit(x);
    

 
void cdq(int l,int r)
    if(l>=r)
        return ;
    int mid=l+r>>1;
    cdq(l,mid);
    cdq(mid+1,r);
    int p=l,q=mid+1,k=l;
    while(p<=mid&&q<=r)
        if(a[p].id<=a[q].id)
            if(a[p].typ==1)
                Add(a[p].y,a[p].val);
            tmp[k++] = a[p++];
        
        else
            if(a[q].typ==2)
                ans[a[q].pos]+=Query(a[q].y)*a[q].val;
            tmp[k++] = a[q++];
        
    
    while(p<=mid)
        if(a[p].typ==1)
            Add(a[p].y,a[p].val);
        tmp[k++]=a[p++];
    
    while(q<=r)
        if(a[q].typ==2)
            ans[a[q].pos]+=Query(a[q].y)*a[q].val;
        tmp[k++]=a[q++];
    
    for(int i=l;i<=r;i++)
        clearr(a[i].y);
        a[i]=tmp[i];
    

 
int main()

    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    std::ios::sync_with_stdio(false);
    int s;
    scanf("%d%d",&s,&n);
    int opt,x,y,val,x1,x2,y1,y2;
    tot=0;
    int m=0;
    while(~scanf("%d",&opt))
        if(opt==3)
            break;
        if(opt==1)
            scanf("%d%d%d",&x,&y,&val);
            a[++m]=node1,x,y,m,val,0;
        
        else
            scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
            ++tot;
            //比x2,y2小的都加上,val赋值1
            a[++m]=node2,x2,y2,m,1,tot;
            //因为他是中间矩阵,而我们加是1~x,1~y的加,所以多出来的要赋值成-1,来减掉,val赋值-1
            a[++m]=node2,x2,y1-1,m,-1,tot;
            a[++m]=node2,x1-1,y2,m,-1,tot;
            //这里多减了一次,要加回来,val赋值1
            a[++m]=node2,x1-1,y1-1,m,1,tot;
 
            ans[tot]+=(x2-x1+1)*(y2-y1+1)*s;  //答案赋初值
        
    
    sort(a+1,a+1+m,cmp);
    cdq(1,m);
    for(int i=1;i<=tot;i++)
        printf("%d\n",ans[i]);
    return 0;

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