jzoj5986. 【WC2019模擬2019.1.4】立體幾何題 (權值線段樹)
阿新 • • 發佈:2019-01-05
題面
題解
不難看出每個點的大小為行列限制中較小的那一個(因為資料保證有解)
對於行的每個限制,能取到的個數是列裡限制大於等於它的數的個數,同理,對於列是行裡大於它的個數(這裡沒有等於,為了避免重複計算)
於是可以對於行列分別開權值線段樹,修改的時候只要把對應的貢獻改一下就好了
//minamoto #include<bits/stdc++.h> #define R register #define ll long long #define fp(i,a,b) for(R int i=a,I=b+1;i<I;++i) #define fd(i,a,b) for(R int i=a,I=b-1;i>I;--i) #define go(u) for(int i=head[u],v=e[i].v;i;i=e[i].nx,v=e[i].v) template<class T>inline bool cmax(T&a,const T&b){return a<b?a=b,1:0;} using namespace std; char buf[1<<21],*p1=buf,*p2=buf; inline char getc(){return p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++;} int read(){ R int res,f=1;R char ch; while((ch=getc())>'9'||ch<'0')(ch=='-')&&(f=-1); for(res=ch-'0';(ch=getc())>='0'&&ch<='9';res=res*10+ch-'0'); return res*f; } char sr[1<<21],z[20];int C=-1,Z=0; inline void Ot(){fwrite(sr,1,C+1,stdout),C=-1;} void print(R ll x){ if(C>1<<20)Ot();if(x<0)sr[++C]='-',x=-x; while(z[++Z]=x%10+48,x/=10); while(sr[++C]=z[Z],--Z);sr[++C]='\n'; } const int N=1e5+5; struct change{int op,pos,x;}c[N]; ll res;int n,m,a[N],b[N],aa[N],bb[N],op,pos,x,lim; struct seg{ struct node{int ls,rs,cnt;ll sum;}t[N<<5]; int rt,tot; void ins(int &p,int l,int r,int x,int ty){ if(!p)p=++tot;t[p].cnt+=ty,t[p].sum+=x*ty; if(l==r)return;int mid=(l+r)>>1; x<=mid?ins(t[p].ls,l,mid,x,ty):ins(t[p].rs,mid+1,r,x,ty); } int q_cnt(int p,int l,int r,int ql,int qr){ if(!p)return 0;if(ql<=l&&qr>=r)return t[p].cnt; int mid=(l+r)>>1,res=0; if(ql<=mid)res+=q_cnt(t[p].ls,l,mid,ql,qr); if(qr>mid)res+=q_cnt(t[p].rs,mid+1,r,ql,qr); return res; } ll q_sum(int p,int l,int r,int ql,int qr){ if(!p)return 0;if(ql<=l&&qr>=r)return t[p].sum; int mid=(l+r)>>1;ll res=0; if(ql<=mid)res+=q_sum(t[p].ls,l,mid,ql,qr); if(qr>mid)res+=q_sum(t[p].rs,mid+1,r,ql,qr); return res; } }A,B; int main(){ freopen("graph.in","r",stdin); freopen("graph.out","w",stdout); n=read(); fp(i,1,n)aa[i]=a[i]=read(),cmax(lim,a[i]); fp(i,1,n)bb[i]=b[i]=read(),cmax(lim,b[i]); m=read(); fp(i,1,m)c[i].op=read(),c[i].pos=read(),c[i].x=read(),cmax(lim,c[i].x); sort(aa+1,aa+1+n),sort(bb+1,bb+1+n); for(R int i=1,j=1;i<=n;++i){ while(j<=n&&aa[i]>bb[j])++j; res+=1ll*aa[i]*(n-j+1); }for(R int i=1,j=1;i<=n;++i){ while(j<=n&&bb[i]>=aa[j])++j; res+=1ll*bb[i]*(n-j+1); }print(res); fp(i,1,n)A.ins(A.rt,0,lim,a[i],1); fp(i,1,n)B.ins(B.rt,0,lim,b[i],1); fp(i,1,m){ op=c[i].op,pos=c[i].pos,x=c[i].x; if(op==0){ res-=1ll*B.q_cnt(B.rt,0,lim,a[pos],lim)*a[pos]; if(a[pos]>0)res-=B.q_sum(B.rt,0,lim,0,a[pos]-1); A.ins(A.rt,0,lim,a[pos],-1); a[pos]=x; A.ins(A.rt,0,lim,a[pos],1); res+=1ll*B.q_cnt(B.rt,0,lim,a[pos],lim)*a[pos]; if(a[pos]>0)res+=B.q_sum(B.rt,0,lim,0,a[pos]-1); }else{ if(b[pos]<lim)res-=1ll*A.q_cnt(A.rt,0,lim,b[pos]+1,lim)*b[pos]; res-=A.q_sum(A.rt,0,lim,0,b[pos]); B.ins(B.rt,0,lim,b[pos],-1); b[pos]=x; B.ins(B.rt,0,lim,b[pos],1); if(b[pos]<lim)res+=1ll*A.q_cnt(A.rt,0,lim,b[pos]+1,lim)*b[pos]; res+=A.q_sum(A.rt,0,lim,0,b[pos]); }print(res); }return Ot(),0; }