bzoj2716

来源：互联网发布：淘宝联盟快速升级编辑：程序博客网时间：2024/06/06 10:40
mdzz数据范围看错了
#include<iostream>#include<algorithm>#include<cstdio>#include<cstdlib>#include<cstring>using namespace std;typedef long long LL;inline int read(){    int x=0;bool f=0;char c=getchar();    for (;c<'0'||c>'9';c=getchar()) f=c=='-'?1:0;    for (;c>='0'&&c<='9';c=getchar()) x=x*10+c-'0';    return f?-x:x;}const int N=500010,M=1000010,oo=0x3f3f3f3f;int n,m,q=0,maxl=0,ans[N],ss[M];struct na{    int x,y,id,t;}e[N<<1],e1[N<<1],e2[N<<1];inline bool cmp(const na &a,const na &b) {    if (a.y!=b.y) return a.y<b.y;    return a.t<b.t;} inline void update1(int x,int a) //维护前缀和{    for (;x<=maxl;x+=x&-x) ss[x]=min(ss[x],a);}inline int query1(int x){    int rec=oo;    for (;x;x-=x&-x) rec=min(rec,ss[x]);    return rec;}inline void recover1(int x){    for (;x<=maxl;x+=x&-x) ss[x]=oo;} inline void update2(int x,int a) //维护后缀和{    for (;x;x-=x&-x) ss[x]=min(ss[x],a);}inline int query2(int x){    int rec=oo;    for (;x<=maxl;x+=x&-x) rec=min(rec,ss[x]);    return rec;}inline void recover2(int x){    for (;x;x-=x&-x) ss[x]=oo;} void solve(int hd,int tl,int l,int r){    if (l>=r||hd>=tl) return;    int mid=l+r>>1;    for (int i=hd;i<=tl;i++)    {        if (e[i].t<=mid&&!e[i].id) update1(e[i].x,-e[i].x-e[i].y);        if (e[i].t>mid&&e[i].id) ans[e[i].id]=min(ans[e[i].id],e[i].x+e[i].y+query1(e[i].x));    }    for (int i=hd;i<=tl;i++)        if (e[i].t<=mid&&!e[i].id) recover1(e[i].x);    for (int i=hd;i<=tl;i++)    {        if (e[i].t<=mid&&!e[i].id) update2(e[i].x,e[i].x-e[i].y);        if (e[i].t>mid&&e[i].id) ans[e[i].id]=min(ans[e[i].id],-e[i].x+e[i].y+query2(e[i].x));    }    for (int i=hd;i<=tl;i++)        if (e[i].t<=mid&&!e[i].id) recover2(e[i].x);    for (int i=tl;i>=hd;i--)    {        if (e[i].t<=mid&&!e[i].id) update1(e[i].x,-e[i].x+e[i].y);        if (e[i].t>mid&&e[i].id) ans[e[i].id]=min(ans[e[i].id],e[i].x-e[i].y+query1(e[i].x));    }    for (int i=tl;i>=hd;i--)        if (e[i].t<=mid&&!e[i].id) recover1(e[i].x);    for (int i=tl;i>=hd;i--)    {        if (e[i].t<=mid&&!e[i].id) update2(e[i].x,e[i].x+e[i].y);        if (e[i].t>mid&&e[i].id) ans[e[i].id]=min(ans[e[i].id],-e[i].x-e[i].y+query2(e[i].x));    }    for (int i=tl;i>=hd;i--)        if (e[i].t<=mid&&!e[i].id) recover2(e[i].x);    int l1=0,l2=0,lp=hd;    for (int i=hd;i<=tl;i++)        if (e[i].t<=mid) e1[++l1]=e[i];            else e2[++l2]=e[i];    for (int i=1;i<=l1;i++) e[lp++]=e1[i];    for (int i=1;i<=l2;i++) e[lp++]=e2[i];    solve(hd,hd+l1-1,l,mid);    solve(hd+l1,tl,mid+1,r);} int main(){    memset(ss,0x3f,sizeof ss);    n=read();m=read();    for (int i=1;i<=n;i++) maxl=max(maxl,e[i].x=read()+1),e[i].y=read()+1,e[i].t=i;    for (int i=n+1;i<=n+m;i++)    {        if (read()==2) e[i].id=++q,ans[q]=oo;        maxl=max(maxl,e[i].x=read()+1),e[i].y=read()+1,e[i].t=i;    }    sort(e+1,e+n+m+1,cmp);    solve(1,n+m,1,n+m);    for (int i=1;i<=q;i++) printf("%d\n",ans[i]);    return 0;}
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