鏈接：https://ac.nowcoder.com/acm/contest/106/H?&headNav=www
來源：牛客網
It’s universally acknowledged that there’re innumerable trees in the campus of HUST.
There is a new leisure ground at the gate of HUST. The president of HUST wants to plant a group of trees on the ground and he wants the trees to be planted exactly like a square matrix. And then the gardens start their work. In the next N days, the gardens can water a tree in coordinate: (x, y) and it makes the height of the tree grows H. Sometimes, the president wants to know how magnificent his trees are. The president will give the gardens two coordinates: (x1, y1), (x2, y2). The gardens should report the sum of the heights of the trees in the sub matrix represents by the two coordinates. There will be A days that the gardens will water a tree. And there will be Q days that the president will ask how magnificent his trees are. Apparently, N=A +Q. Furthermore, all the trees’ heights are 0 in the very beginning.
All the absolute values of the inputs and the results are all nonnegative and won't exceed 231-1.
輸入描述:
The first line consists of two integers N and W(0≤A≤105,0≤Q≤105,1≤W≤105)(0 \leq A \leq 10^{5},0 \leq Q \leq 10^{5}, 1\leq W\leq 10^{5})(0≤A≤105,0≤Q≤105,1≤W≤105). N represents the number of days and W represents the size of the matrix. Followed by N lines in two forms:
1 x y H
2 x1 y1 x2 y2
The first one represents the gardeners will water the tree in coordinate (x, y) and it will grow H(1≤x,y≤W)(1\leq x,y\leq W)(1≤x,y≤W).
The second one represents that you need to output the sum of the heights of the trees in the sub matrix represents by the two coordinates.
輸出描述:
Your output should consist of Q lines. Each line contains an integer, the result of a query.
示例1
輸入
4 5 1 1 3 2 2 1 1 2 2 1 2 2 3 2 1 1 3 3
輸出
0 5
題意：給你一個矩陣，n個操作：
1 x y h 將(x,y)中的數字加上h
2 x1 y1 x2 y2 求(x1,y1)到(x2,y2)這個子矩陣內所有數字的和。
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn=1e5+100;
const int maxm=5e5+100;
int cnt,n,w,tree[maxn],ans[maxn];
int lowbit(int x)
{
return x&-x;
}
void update(int tar,int val)
{
for(int i=tar;i<=w;i+=lowbit(i))
tree[i]+=val;
}
int query(int tar)
{
int res=0;
for(int i=tar;i;i-=lowbit(i)) res+=tree[i];
return res;
}
struct node{
int x;
int y;
int h;
int flag;
bool operator < (const node &t)const
{
if(x<t.x) return 1;
if(x>t.x) return 0;
if(y<t.y) return 1;
if(y>t.y) return 0;
return flag<t.flag;
}
}a[maxm],tmp[maxm];
void CDQ(int l,int r)
{
if(l==r) return ;
int mid=(l+r)>>1,p=l,q=mid+1,cont=l-1;
CDQ(l,mid);CDQ(mid+1,r);
while(p<=mid&&q<=r)
{
if(a[p]<a[q])
{
if(a[p].flag==1) update(a[p].y,a[p].h);
tmp[++cont]=a[p++];
}
else
{
if(a[q].flag==2) ans[a[q].h]+=query(a[q].y);
if(a[q].flag==3) ans[a[q].h]-=query(a[q].y);
tmp[++cont]=a[q++];
}
}
while(q<=r)
{
if(a[q].flag==2) ans[a[q].h]+=query(a[q].y);
if(a[q].flag==3) ans[a[q].h]-=query(a[q].y);
tmp[++cont]=a[q++];
}
for(int i=l;i<p;i++) if(a[i].flag==1) update(a[i].y,-a[i].h);
while(p<=mid) tmp[++cont]=a[p++];
for(int i=l;i<=r;i++) a[i]=tmp[i];
}
int main()
{
int op,x1,x2,y1,y2,ct;
cnt=ct=0;
scanf("%d%d",&n,&w);
while(n--)
{
scanf("%d",&op);
if(op==1)
{
scanf("%d%d%d",&x1,&y1,&y2);
a[++cnt].x=x1;a[cnt].y=y1;a[cnt].h=y2;
a[cnt].flag=1;
}
else
{
scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
a[++cnt].x=x2;a[cnt].y=y2;a[cnt].h=++ct;a[cnt].flag=2;
a[++cnt].x=x1-1;a[cnt].y=y1-1;a[cnt].h=ct;a[cnt].flag=2;
a[++cnt].x=x1-1;a[cnt].y=y2;a[cnt].h=ct;a[cnt].flag=3;
a[++cnt].x=x2;a[cnt].y=y1-1;a[cnt].h=ct;a[cnt].flag=3;
}
}
CDQ(1,cnt);
for(int i=1;i<=ct;i++) printf("%d\n",ans[i]);
return 0;
}