BZOJ 2683 简单题 CDQ分治+树状数组

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题目大意：维护一个矩阵，单点修改，子矩阵查询，不强制在线

CDQ分治裸题。。。逗我。。。

同BZOJ 1176 Mokia 题解见 http://blog.csdn.net/popoqqq/article/details/39672705

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#define M 500500using namespace std;struct Query{int x,y,type;//type>0 修改//type==-1 查询 正//type==-3 查询 负Query() {}Query(int _,int __,int ___):x(_),y(__),type(___) {}}mempool[800800],*q[800800],*nq[800800];int n,m,ans[800800];int stack[200200],top;bool Compare(Query *q1,Query *q2){return q1->x < q2->x;}namespace BIT{int c[M],tim[M],T;inline void Update(int x,int y){for(;x<=n;x+=x&-x){if(tim[x]!=T)tim[x]=T,c[x]=0;c[x]+=y;}}inline int Get_Ans(int x){int re=0;for(;x;x-=x&-x)if(tim[x]==T)re+=c[x];return re;}}void CDQ_Divide_And_Conquer(int l,int r){using namespace BIT;int i,j,mid=l+r>>1;if(l==r) return ;int l1=l,l2=mid+1;for(i=l;i<=r;i++)if(q[i]-mempool<=mid)nq[l1++]=q[i];else nq[l2++]=q[i];memcpy(q+l,nq+l,sizeof(q[0])*(r-l+1) );CDQ_Divide_And_Conquer(l,mid);for(++T,j=l,i=mid+1;i<=r;i++){for(;j<=mid&&q[j]->x<=q[i]->x;j++)if(q[j]->type>0)Update(q[j]->y,q[j]->type);if(q[i]->type<0)ans[q[i]-mempool]+=Get_Ans(q[i]->y)*(q[i]->type+2);}CDQ_Divide_And_Conquer(mid+1,r);l1=l,l2=mid+1;for(i=l;i<=r;i++){if(l2>r||l1<=mid&&Compare(q[l1],q[l2]))nq[i]=q[l1++];else nq[i]=q[l2++];}memcpy(q+l,nq+l,sizeof(q[0])*(r-l+1) );}int main(){int i,j,x,y,z,x1,y1,x2,y2,p;cin>>n;while(1){scanf("%d",&p);if(p==1){scanf("%d%d%d",&x,&y,&z);++m,q[m]=&(mempool[m]=Query(x,y,z));}else if(p==2){scanf("%d%d%d%d",&x1,&y1,&x2,&y2);stack[++top]=m;++m,q[m]=&(mempool[m]=Query(x1-1,y1-1,-1));++m,q[m]=&(mempool[m]=Query(x1-1,y2,-3));++m,q[m]=&(mempool[m]=Query(x2,y1-1,-3));++m,q[m]=&(mempool[m]=Query(x2,y2,-1));}else break;}sort(q+1,q+m+1,Compare);CDQ_Divide_And_Conquer(1,m);for(i=1;i<=top;i++){int temp=0;for(j=1;j<=4;j++)temp+=ans[stack[i]+j];printf("%d\n",temp);}return 0;}

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