洛谷 P3332 [ZJOI2013]K大数查询 || bzoj3110

  1 #include<cstdio>
  2 #include<algorithm>
  3 #include<cstring>
  4 using namespace std;
  5 typedef long long LL;
  6 struct Q
  7 {
  8     LL type,a,b,c,num;
  9 }q[50100],qt1[50100],qt2[50100];
 10 LL ans[50100],qnum;
 11 LL n,m;
 12 namespace SegT
 13 {
 14 #define mid (l+((r-l)>>1))
 15 #define lc (num<<1)
 16 #define rc (num<<1|1)
 17 LL dat[200100],addv[200100];
 18 LL L,R,x;
 19 void pd(LL l,LL r,LL num)
 20 {
 21     if(addv[num])
 22     {
 23         dat[lc]+=(mid-l+1)*addv[num];addv[lc]+=addv[num];
 24         dat[rc]+=(r-mid)*addv[num];addv[rc]+=addv[num];
 25         addv[num]=0;
 26     }
 27 }
 28 void _addx(LL l,LL r,LL num)
 29 {
 30     if(L<=l&&r<=R)    {dat[num]+=(r-l+1)*x;addv[num]+=x;return;}
 31     pd(l,r,num);
 32     if(L<=mid)    _addx(l,mid,lc);
 33     if(mid<R)    _addx(mid+1,r,rc);
 34     dat[num]=dat[lc]+dat[rc];
 35 }
 36 LL _query(LL l,LL r,LL num)
 37 {
 38     if(L<=l&&r<=R)    return dat[num];
 39     pd(l,r,num);LL ans=0;
 40     if(L<=mid)    ans+=_query(l,mid,lc);
 41     if(mid<R)    ans+=_query(mid+1,r,rc);
 42     return ans;
 43 }
 44 void addx(LL l,LL r,LL d)
 45 {
 46     L=l;R=r;x=d;_addx(1,n,1);
 47 }
 48 LL query(LL l,LL r)
 49 {
 50     L=l;R=r;return _query(1,n,1);
 51 }
 52 #undef mid
 53 #undef lc
 54 #undef rc
 55 }
 56 void solve(LL lp,LL rp,LL l,LL r)
 57 {
 58     if(lp>rp)    return;
 59     LL i;
 60     if(l==r)
 61     {
 62         for(i=lp;i<=rp;i++)
 63             if(q[i].type==2)
 64                 ans[q[i].num]=l;
 65         return;
 66     }
 67     LL mid=l+((r-l)>>1),tlen1=0,tlen2=0,t;
 68     for(i=lp;i<=rp;i++)
 69     {
 70         if(q[i].type==1)
 71         {
 72             if(q[i].c>mid)
 73             {
 74                 SegT::addx(q[i].a,q[i].b,1);
 75                 qt1[++tlen1]=q[i];
 76             }
 77             else
 78                 qt2[++tlen2]=q[i];
 79         }
 80         else
 81         {
 82             t=SegT::query(q[i].a,q[i].b);
 83             if(t>=q[i].c)
 84                 qt1[++tlen1]=q[i];
 85             else
 86                 qt2[++tlen2]=q[i],qt2[tlen2].c-=t;
 87         }
 88     }
 89     for(i=lp;i<=rp;i++)
 90         if(q[i].type==1&&q[i].c>mid)
 91             SegT::addx(q[i].a,q[i].b,-1);
 92     memcpy(q+lp,qt1+1,sizeof(Q)*tlen1);
 93     memcpy(q+lp+tlen1,qt2+1,sizeof(Q)*tlen2);
 94     solve(lp,lp+tlen1-1,mid+1,r);
 95     solve(lp+tlen1,rp,l,mid);
 96 }
 97 int main()
 98 {
 99     LL i;
100     scanf("%lld%lld",&n,&m);
101     for(i=1;i<=m;i++)
102     {
103         scanf("%lld%lld%lld%lld",&q[i].type,&q[i].a,&q[i].b,&q[i].c);
104         if(q[i].type==2)    q[i].num=++qnum;
105     }
106     solve(1,m,-50000,50000);
107     for(i=1;i<=qnum;i++)    printf("%lld\n",ans[i]);
108     return 0;
109 }