[BOI2007]Mokia 摩基亚
我才不会告诉你我cmp写错了,然后调了一年
CDQ分治模板题
一维时间,二三维x,y坐标
避免树状数组下标出现0炸掉,把跟长度有关的都+1
答案用容斥原理即可(类似于二维前缀和?)
1 //LevenKoko 2 #include<bits/stdc++.h> 3 #define lowbit(x) (x&(-x)) 4 using namespace std; 5 inline int read(){ 6 int ans=0,f=1;char chr=getchar(); 7 while(!isdigit(chr)){if(chr=='-')f=-1;chr=getchar();} 8 while(isdigit(chr)) {ans=(ans<<3)+(ans<<1)+chr-48;chr=getchar();} 9 return ans*f; 10 }const int M=2e6+5,N=2e5+5; 11 int opt,s[M],n,t,x,y,z,xx,yy; 12 struct P{int t,x,y,cnt,id;}a[N]; 13 bool cmp1(P x,P y){if(x.x==y.x) return x.y<y.y;return x.x<y.x;} 14 bool cmp2(P x,P y){return x.t<y.t;} 15 inline void Add(int x,int z){for(;x<=n;x+=lowbit(x)) s[x]+=z;} 16 inline int Ask(int x){int ans=0;for(;x;x-=lowbit(x)) ans+=s[x];return ans;} 17 inline void Plu(int x,int y){a[++t]=(P){t,x,y,0,1};} 18 inline void Init(){ 19 read(),n=read()+1; 20 while(opt=read()){ 21 if(opt==1)x=read()+1,y=read()+1,z=read(),a[++t]=(P){t,x,y,z,0}; 22 if(opt==2)x=read(),y=read(),xx=read()+1,yy=read()+1,Plu(x,y),Plu(xx,yy),Plu(xx,y),Plu(x,yy); 23 if(opt==3) break; 24 } 25 } 26 inline void Solve(int l,int r){ 27 if(l==r) return; 28 int mid=l+r>>1; 29 Solve(l,mid),Solve(mid+1,r); 30 sort(a+l,a+mid+1,cmp1),sort(a+mid+1,a+r+1,cmp1); 31 int i=l,j=mid+1; 32 for(;j<=r;j++){ 33 while(a[i].x<=a[j].x&&i<=mid){ 34 if(a[i].id==0) Add(a[i].y,a[i].cnt); 35 i++; 36 }if(a[j].id==1) a[j].cnt+=Ask(a[j].y); 37 } 38 for(j=l;j<i;j++) if(a[j].id==0)Add(a[j].y,-a[j].cnt); 39 } 40 inline void Get(){ 41 sort(a+1,a+t+1,cmp2); 42 for(int i=1;i<=t;i++) 43 if(a[i].id==1) 44 printf("%d\n",a[i].cnt+a[i+1].cnt-a[i+2].cnt-a[i+3].cnt),i+=3; 45 } 46 int main(){ 47 return Init(),Solve(1,t),Get(),0; 48 }