題目鏈接：點擊查看

題目大意：給出一個二維平面坐標系，需要執(zhí)行數(shù)次操作，具體操作分為下列兩種：

1 x y a：坐標 $(x,y)$ 加上 $a$ 個點

2 x₁ y₁ x₂ y₂：查詢以 $(x_1,y_1)$ 為左下角、$(x_2,y_2)$ 為右上角的矩陣中有多少個點

題目分析：三種做法，但是樹套樹內(nèi)存不太夠，所以拿不了滿分。

剩下的四叉樹跑的巨慢，cdq分治表現(xiàn)還算不錯。因為寫 $cdq$ 的時候詢問和加點并不會沖突，所以不需要去重，只需要保證在維度相同的時候，令加點操作在詢問操作之前即可

代碼：
cdq分治

// Problem: P4390 [BOI2007]Mokia 摩基亞 // Contest: Luogu // URL: https://www.luogu.com.cn/problem/P4390 // Memory Limit: 125 MB // Time Limit: 2000 ms // // Powered by CP Editor (https://cpeditor.org)// #pragma GCC optimize(2) // #pragma GCC optimize("Ofast","inline","-ffast-math") // #pragma GCC target("avx,sse2,sse3,sse4,mmx") #include<iostream> #include<cstdio> #include<string> #include<ctime> #include<cmath> #include<cstring> #include<algorithm> #include<stack> #include<climits> #include<queue> #include<map> #include<set> #include<sstream> #include<cassert> #include<bitset> #include<list> #include<unordered_map> #define lowbit(x) (x&-x) using namespace std; typedef long long LL; typedef unsigned long long ull; template<typename T> inline void read(T &x) {T f=1;x=0;char ch=getchar();while(0==isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}while(0!=isdigit(ch)) x=(x<<1)+(x<<3)+ch-'0',ch=getchar();x*=f; } template<typename T> inline void write(T x) {if(x<0){x=~(x-1);putchar('-');}if(x>9)write(x/10);putchar(x%10+'0'); } const int inf=0x3f3f3f3f; const int N=1e6+100; struct Node {int a,b,c,type,sgn,id,val;bool operator<(const Node& t)const {if(a!=t.a) return a<t.a;if(b!=t.b) return b<t.b;if(c!=t.c) return c<t.c;return type<t.type;} }a[N],t[N],temp[N]; int c[N],ans[N]; void add(int x,int val) {for(int i=x;i<N;i+=lowbit(i)) c[i]+=val; } int ask(int x) {int ans=0;for(int i=x;i>0;i-=lowbit(i)) ans+=c[i];return ans; } 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,tot=l;while(p<=mid&&q<=r) {if(a[p].b<=a[q].b) {if(a[p].type==0) {//addadd(a[p].c,a[p].val);}t[tot++]=a[p++];} else {if(a[q].type==1) {//askans[a[q].id]+=a[q].sgn*ask(a[q].c);}t[tot++]=a[q++];}}while(p<=mid) {if(a[p].type==0) {add(a[p].c,a[p].val);}t[tot++]=a[p++];}while(q<=r) {if(a[q].type==1) {ans[a[q].id]+=a[q].sgn*ask(a[q].c);}t[tot++]=a[q++];}for(int i=l;i<=mid;i++) {if(a[i].type==0) {add(a[i].c,-a[i].val);}}for(int i=l;i<=r;i++) {a[i]=t[i];} } int main() { #ifndef ONLINE_JUDGE // freopen("data.in.txt","r",stdin); // freopen("data.out.txt","w",stdout); #endif // ios::sync_with_stdio(false);int op,qcnt=0,n=0,t=0;while(scanf("%d",&op)!=EOF&&op!=3) {t++;if(op==0) {read(op);} else if(op==1) {int x,y,val;read(x),read(y),read(val);a[++n]={x,y,t,0,0,0,val};} else if(op==2) {int x1,y1,x2,y2;read(x1),read(y1),read(x2),read(y2);qcnt++;a[++n]={x2,y2,t,1,1,qcnt,-1};a[++n]={x1-1,y1-1,t,1,1,qcnt,-1};a[++n]={x1-1,y2,t,1,-1,qcnt,-1};a[++n]={x2,y1-1,t,1,-1,qcnt,-1};}}sort(a+1,a+1+n);CDQ(1,n);for(int i=1;i<=qcnt;i++) {printf("%d\n",ans[i]);}return 0; }

四叉樹：

// Problem: P4390 [BOI2007]Mokia 摩基亞 // Contest: Luogu // URL: https://www.luogu.com.cn/problem/P4390 // Memory Limit: 125 MB // Time Limit: 2000 ms // // Powered by CP Editor (https://cpeditor.org)// #pragma GCC optimize(2) // #pragma GCC optimize("Ofast","inline","-ffast-math") // #pragma GCC target("avx,sse2,sse3,sse4,mmx") #include<iostream> #include<cstdio> #include<string> #include<ctime> #include<cmath> #include<cstring> #include<algorithm> #include<stack> #include<climits> #include<queue> #include<map> #include<set> #include<sstream> #include<cassert> #include<bitset> #include<list> #include<unordered_map> #define lowbit(x) (x&-x) using namespace std; typedef long long LL; typedef unsigned long long ull; template<typename T> inline void read(T &x) {T f=1;x=0;char ch=getchar();while(0==isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}while(0!=isdigit(ch)) x=(x<<1)+(x<<3)+ch-'0',ch=getchar();x*=f; } template<typename T> inline void write(T x) {if(x<0){x=~(x-1);putchar('-');}if(x>9)write(x/10);putchar(x%10+'0'); } const int inf=0x3f3f3f3f; const int N=160005; int ld[N*22],lu[N*22],rd[N*22],ru[N*22],sum[N*22],cnt,rt,UP; int newnode() {cnt++;ld[cnt]=lu[cnt]=rd[cnt]=ru[cnt]=sum[cnt]=0;return cnt; } void update(int &k,int x,int y,int val,int XL=1,int XR=UP,int YL=1,int YR=UP) {if(!k) k=newnode();sum[k]+=val;if(XL==XR&&YL==YR) return;int midx=(XL+XR)>>1,midy=(YL+YR)>>1;if(midx>=x) {if(midy>=y) update(ld[k],x,y,val,XL,midx,YL,midy);else if(midy!=YR) update(lu[k],x,y,val,XL,midx,midy+1,YR);} else if(midx!=XR) {if(midy>=y) update(rd[k],x,y,val,midx+1,XR,YL,midy);else if(midy!=YR) update(ru[k],x,y,val,midx+1,XR,midy+1,YR);} } int query(int k,int xl,int xr,int yl,int yr,int XL=1,int XR=UP,int YL=1,int YR=UP) {if(XL>=xl&&XR<=xr&&YL>=yl&&YR<=yr) return sum[k];int midx=(XL+XR)>>1,midy=(YL+YR)>>1,ans=0;if(midx>=xl && midy>=yl && ld[k]) ans+=query(ld[k],xl,xr,yl,yr,XL,midx,YL,midy);if(midx>=xl && midy!=YR && lu[k]) ans+=query(lu[k],xl,xr,yl,yr,XL,midx,midy+1,YR);if(midx<xr && midx!=XR && midy>=yl && rd[k]) ans+=query(rd[k],xl,xr,yl,yr,midx+1,XR,YL,midy);if(midx<xr && midx!=XR && midy<yr && midy!=YR && ru[k]) ans+=query(ru[k],xl,xr,yl,yr,midx+1,XR,midy+1,YR);return ans; } void init() {cnt=0;rt=newnode(); } int main() { #ifndef ONLINE_JUDGE // freopen("data.in.txt","r",stdin); // freopen("data.out.txt","w",stdout); #endif // ios::sync_with_stdio(false);init();int op;while(scanf("%d",&op)!=EOF&&op!=3) {if(op==0) {read(UP);} else if(op==1) {int x,y,a;read(x),read(y),read(a);update(rt,x,y,a);} else if(op==2) {int x1,y1,x2,y2;read(x1),read(y1),read(x2),read(y2);printf("%d\n",query(rt,x1,x2,y1,y2));}}return 0; }

樹套樹：

// Problem: P4390 [BOI2007]Mokia 摩基亞 // Contest: Luogu // URL: https://www.luogu.com.cn/problem/P4390 // Memory Limit: 125 MB // Time Limit: 2000 ms // // Powered by CP Editor (https://cpeditor.org)// #pragma GCC optimize(2) // #pragma GCC optimize("Ofast","inline","-ffast-math") // #pragma GCC target("avx,sse2,sse3,sse4,mmx") #include<iostream> #include<cstdio> #include<string> #include<ctime> #include<cmath> #include<cstring> #include<algorithm> #include<stack> #include<climits> #include<queue> #include<map> #include<set> #include<sstream> #include<cassert> #include<bitset> #include<list> #include<unordered_map> #define lowbit(x) (x&-x) using namespace std; typedef long long LL; typedef unsigned long long ull; template<typename T> inline void read(T &x) {T f=1;x=0;char ch=getchar();while(0==isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}while(0!=isdigit(ch)) x=(x<<1)+(x<<3)+ch-'0',ch=getchar();x*=f; } template<typename T> inline void write(T x) {if(x<0){x=~(x-1);putchar('-');}if(x>9)write(x/10);putchar(x%10+'0'); } const int inf=0x3f3f3f3f; const int N=2e6+100; int cnt,UP; struct Seg {struct Node {int l,r;int sum;}tree[160010*21*21];int newnode() {cnt++;tree[cnt].l=tree[cnt].r=0;tree[cnt].sum=0;return cnt;}void update(int &k,int l,int r,int pos,int val) {if(!k) {k=newnode();}tree[k].sum+=val;if(l==r) {return;}int mid=(l+r)>>1;if(pos<=mid) {update(tree[k].l,l,mid,pos,val);} else {update(tree[k].r,mid+1,r,pos,val);}}int query(int k,int l,int r,int ql,int qr) {if(!k||l>qr||r<ql) {return 0;}if(l>=ql&&r<=qr) {return tree[k].sum;}int mid=(l+r)>>1;return query(tree[k].l,l,mid,ql,qr)+query(tree[k].r,mid+1,r,ql,qr);} }SEG; struct Bit {int root[N];void add(int x,int y,int val) {for(int i=x;i<N;i+=lowbit(i)) {SEG.update(root[i],1,UP,y,val);}}int ask(int x,int y1,int y2) {int ans=0;for(int i=x;i>0;i-=lowbit(i)) {ans+=SEG.query(root[i],1,UP,y1,y2);}return ans;} }BIT; void init() {cnt=-1;SEG.newnode();memset(BIT.root,0,sizeof(BIT.root)); } int main() { #ifndef ONLINE_JUDGE // freopen("data.in.txt","r",stdin); // freopen("data.out.txt","w",stdout); #endif // ios::sync_with_stdio(false);init();scanf("%*d%d",&UP);int op;while(scanf("%d",&op)!=EOF&&op!=3) {if(op==1) {int x,y,a;read(x),read(y),read(a);BIT.add(x,y,a);} else if(op==2) {int x1,y1,x2,y2;read(x1),read(y1),read(x2),read(y2);printf("%d\n",BIT.ask(x2,y1,y2)-BIT.ask(x1-1,y1,y2));}}return 0; }

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