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题意分析

首先\(\%\%\%\%olinr\)以及花_Q\(julao\)当场切题

然后就是怎么求

\[max(|a-A|,|b-B|)=max(a-A,A-a,B-b,b-B)\]

我们令\(x_1=(a+b)/2,x_2=(A+B)/2,y_1=(a-b)/2,y_2=(A-B)/2\)

\[max(x_1+y_1-x_2-y_2,x_2+y_2-x_1-y_1,x_1-y_1-x_2+y_2,x_2-y_2-x_1+y_1)\]

\[=max(x_1-x_2,x_2-x_1)+max(y_1-y_2,y_2-y_1)\]

\[=|x_1-x_2|+|y_1-y_2|\]

那么就是用主席树维护\(\{A_i+B_i\}\)以及\(\{A_i-B_i\}\)的中位数

CODE:

/*-------------OI使我快乐-------------*/
int n,m,tot;
int cdy[M],wzy[M],lc[M],rc[M];
int root[M][2],siz[M*25][2],lson[M*25][2],rson[M*25][2];
ll sum[M*25][2];
IL void insert(int A,int &B,int le,int ri,int pos,int knd)
{
//  printf("insert %d %d %d\n",le,ri,pos);
    B=++tot;
    siz[B][knd]=siz[A][knd]+1;
    sum[B][knd]=sum[A][knd]+pos;
    if(le==ri) return;
    int mid=(le+(ll)ri)>>1;
    if(pos<=mid) insert(lson[A][knd],lson[B][knd],le,mid,pos,knd),rson[B][knd]=rson[A][knd];
    else insert(rson[A][knd],rson[B][knd],mid+1,ri,pos,knd),lson[B][knd]=lson[A][knd];
}
IL int qury_kth(int A,int B,int le,int ri,int k,int knd)
{
    if(le==ri) return le;
    int mid=(le+(ll)ri)>>1,tmp=siz[lson[B][knd]][knd]-siz[lson[A][knd]][knd];
    if(k<=tmp) return qury_kth(lson[A][knd],lson[B][knd],le,mid,k,knd);
    else return qury_kth(rson[A][knd],rson[B][knd],mid+1,ri,k-tmp,knd);
}
IL ll qury_sum(int A,int B,int lenow,int rinow,int le,int ri,int knd)
{
    if(le<=lenow&&rinow<=ri) return sum[B][knd]-sum[A][knd];
    int mid=(lenow+(ll)rinow)>>1;ll res=0;
    if(le<=mid) res+=qury_sum(lson[A][knd],lson[B][knd],lenow,mid,le,ri,knd);
    if(mid<ri) res+=qury_sum(rson[A][knd],rson[B][knd],mid+1,rinow,le,ri,knd);
    return res;
}
IL int qury_cnt(int A,int B,int lenow,int rinow,int le,int ri,int knd)
{
    if(le<=lenow&&rinow<=ri) return siz[B][knd]-siz[A][knd];
    int mid=(lenow+(ll)rinow)>>1,res=0;
    if(le<=mid) res+=qury_cnt(lson[A][knd],lson[B][knd],lenow,mid,le,ri,knd);
    if(mid<ri) res+=qury_cnt(rson[A][knd],rson[B][knd],mid+1,rinow,le,ri,knd);
    return res;
}
int main()
{
    freopen("spy.in","r",stdin);
    freopen("spy.out","w",stdout);
    read(n);read(m);
    for(R int i=1;i<=n;++i) read(cdy[i]);
    for(R int i=1;i<=n;++i) read(wzy[i]);
    for(R int i=1;i<=n;++i) lc[i]=cdy[i]+wzy[i],rc[i]=cdy[i]-wzy[i];
    for(R int i=1;i<=n;++i) insert(root[i-1][0],root[i][0],-maxn,maxn,lc[i],0);
    tot=0;
    for(R int i=1;i<=n;++i) insert(root[i-1][1],root[i][1],-maxn,maxn,rc[i],1);
    while(m--)
    {
        int le,ri,midx,midy;ll lx,rx,ly,ry;
        read(le);read(ri);
        midx=qury_kth(root[le-1][0],root[ri][0],-maxn,maxn,(ri-le+2)/2,0);
//      puts("now now now");
        midy=qury_kth(root[le-1][1],root[ri][1],-maxn,maxn,(ri-le+2)/2,1);
//      puts("now now now");
//      printf("%d %d\n",midx,midy);
        lx=1ll*qury_cnt(root[le-1][0],root[ri][0],-maxn,maxn,-maxn,midx,0)*midx-qury_sum(root[le-1][0],root[ri][0],-maxn,maxn,-maxn,midx,0);
        rx=qury_sum(root[le-1][0],root[ri][0],-maxn,maxn,midx,maxn,0)-1ll*qury_cnt(root[le-1][0],root[ri][0],-maxn,maxn,midx,maxn,0)*midx;
//      printf("nmow %lld %lld\n",lx,rx);
        ly=1ll*qury_cnt(root[le-1][1],root[ri][1],-maxn,maxn,-maxn,midy,1)*midy-qury_sum(root[le-1][1],root[ri][1],-maxn,maxn,-maxn,midy,1);
        ry=qury_sum(root[le-1][1],root[ri][1],-maxn,maxn,midy,maxn,1)-1ll*qury_cnt(root[le-1][1],root[ri][1],-maxn,maxn,midy,maxn,1)*midy;
        printf("%.2f\n",(lx+rx+ly+ry)/2.0);     
    }
    fclose(stdin);
    fclose(stdout);
    return 0;
}

HEOI 2019 RP++

考试题 T3

标签：主席树   int   a+b   isp   insert   play   img   res   open   

原文地址：https://www.cnblogs.com/LovToLZX/p/10623675.html