二维树状数组的区间加减及查询 tyvj 1716 上帝造题的七分钟

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <queue>
#include <cmath>
#include <stack>
#include <map>

#pragma comment(linker, "/STACK:1024000000");
#define EPS (1e-8)
#define LL long long
#define ULL unsigned long long
#define _LL __int64
#define _INF 0x3f3f3f3f
#define Mod 9999991
#define lowbit(x) (x&(-x))

using namespace std;

const int N = 2049;

int a

;
int ax

;
int ay

;
int axy

;

void Updata(int a

,int x,int y,int n,int m,int data)
{
for(; x <= n; x += lowbit(x))
{
for(int t = y; t <= m; t += lowbit(t))
{
a[x][t] += data;
}
}
}

int Query(int a

,int x,int y)
{
int sum = 0;

while(x > 0)
{
for(int t = y; t > 0; t -= lowbit(t))
{
sum += a[x][t];
}
x -= lowbit(x);
}

return sum;
}

int sum(int x,int y)
{
return Query(a,x,y)*(x+1)*(y+1) - (y+1)*Query(ax,x,y) - (x+1)*Query(ay,x,y) + Query(axy,x,y);
}

void Add(int x1,int y1,int x2,int y2,int n,int m,int data)
{
Updata(a,x1,y1,n,m,data);
Updata(a,x2+1,y1,n,m,-data);
Updata(a,x1,y2+1,n,m,-data);
Updata(a,x2+1,y2+1,n,m,data);

Updata(ax,x1,y1,n,m,x1*data);
Updata(ax,x2+1,y1,n,m,-(x2+1)*data);
Updata(ax,x1,y2+1,n,m,-x1*data);
Updata(ax,x2+1,y2+1,n,m,(x2+1)*data);

Updata(ay,x1,y1,n,m,y1*data);
Updata(ay,x2+1,y1,n,m,-y1*data);
Updata(ay,x1,y2+1,n,m,-(y2+1)*data);
Updata(ay,x2+1,y2+1,n,m,(y2+1)*data);

Updata(axy,x1,y1,n,m,x1*y1*data);
Updata(axy,x2+1,y1,n,m,-(x2+1)*y1*data);
Updata(axy,x1,y2+1,n,m,-x1*(y2+1)*data);
Updata(axy,x2+1,y2+1,n,m,(x2+1)*(y2+1)*data);
}

int main()
{
int n,m;

char order[10];

int x1,x2,y1,y2,data;

memset(a,0,sizeof(a));
memset(ax,0,sizeof(ax));
memset(ay,0,sizeof(ay));
memset(axy,0,sizeof(axy));

while(scanf("%s",order) != EOF)
{
if(order[0] == 'L')
{
scanf("%d %d %d %d %d",&x1,&y1,&x2,&y2,&data);

Add(x1,y1,x2,y2,n,m,data);
}
else if(order[0] == 'k')
{
scanf("%d %d %d %d",&x1,&y1,&x2,&y2);

printf("%d\n",sum(x2,y2) - sum(x1-1,y2) - sum(x2,y1-1) + sum(x1-1,y1-1));
}
else
{
scanf("%d %d",&n,&m);
}
}

return 0;
}