bzoj2234 Berhatton

每个点覆盖的区域是一个斜放的正方形，用一条x+y=b的扫描线扫一遍，需要维护区间覆盖和单点查询。线段树就可以了。

#include<cstdio>#include<algorithm>using namespace std;#define LL long longconst int maxn=100010;int rd(){    int x=0;    char c=getchar();    while (c<'0'||c>'9') c=getchar();    while (c>='0'&&c<='9')    {        x=x*10+c-'0';        c=getchar();    }    return x;}struct str{    int k,l,r,id;    LL p;    bool operator < (const str &s) const    {        if (p!=s.p) return p<s.p;        return k>s.k;    }}a[maxn*3];int xx[maxn],yy[maxn],dis[maxn],ord[5*maxn],tag[15*maxn],ans[maxn],n,m,k,tot,res;void modi(int u,int L,int R,int l,int r,int x){    if (l<=L&&R<=r)    {        tag[u]+=x;        return;    }    int mid=(L+R)/2;    if (l<=mid) modi(u*2,L,mid,l,r,x);    if (r>mid) modi(u*2+1,mid+1,R,l,r,x);}int qry(int u,int L,int R,int p){    if (L==R) return tag[u];    int mid=(L+R)/2;    if (p<=mid) return qry(u*2,L,mid,p)+tag[u];    return qry(u*2+1,mid+1,R,p)+tag[u];}int main(){    /*freopen("berhatton.in","r",stdin);    freopen("berhatton.out","w",stdout);*/    int x,l,r;    n=rd();    k=rd();    for (int i=1;i<=n;i++)    {        xx[i]=rd();        yy[i]=rd();        dis[i]=rd();         ord[++m]=xx[i]-yy[i];        ord[++m]=xx[i]+dis[i]-yy[i];        ord[++m]=xx[i]-dis[i]-yy[i];    }    sort(ord+1,ord+m+1);    m=unique(ord+1,ord+m+1)-ord-1;    for (int i=1;i<=n;i++)    {        x=lower_bound(ord+1,ord+m+1,xx[i]-yy[i])-ord;        a[++tot]=(str){0,x,x,i,xx[i]+yy[i]};        l=lower_bound(ord+1,ord+m+1,xx[i]-yy[i]-dis[i])-ord;        r=lower_bound(ord+1,ord+m+1,xx[i]-yy[i]+dis[i])-ord;        a[++tot]=(str){1,l,r,0,xx[i]+yy[i]-dis[i]};        a[++tot]=(str){-1,l,r,0,(LL)xx[i]+yy[i]+dis[i]};    }    sort(a+1,a+tot+1);    for (int i=1;i<=tot;i++)        if (a[i].k) modi(1,1,m,a[i].l,a[i].r,a[i].k);        else if (qry(1,1,m,a[i].l)>k) ans[++res]=a[i].id;    sort(ans+1,ans+res+1);    printf("%d\n",res);    for (int i=1;i<=res;i++) printf("%d%c",ans[i],i==res?'\n':' ');    //fclose(stdout);}