bzoj 3110 K大数查询

  1 #include<iostream>
  2 #include<cstdio>
  3 #include<cstring>
  4 #include<algorithm>
  5 #define int long long
  6 #define N 50005
  7 #define ls x*2,l,mid
  8 #define rs x*2+1,mid+1,r
  9 using namespace std;
 10 int n,m;
 11 struct node
 12 {
 13     int sign,x,y,c;
 14 }op[N];
 15 int q[N],ans[N];
 16 int a[N*8],lazy[N*8];
 17 void push_down(int x,int l,int mid,int r)
 18 {
 19     if(lazy[x]!=0)
 20     {
 21         lazy[x*2]+=lazy[x];
 22         lazy[x*2+1]+=lazy[x];
 23         a[x*2]+=lazy[x]*(mid-l+1);
 24         a[x*2+1]+=lazy[x]*(r-mid);
 25         lazy[x]=0;
 26     }
 27     return ;
 28 }
 29 int qur(int x,int l,int r,int ll,int rr)
 30 {
 31     if(l>=ll&&r<=rr)return a[x];
 32     int mid=(l+r)>>1;
 33     push_down(x,l,mid,r);
 34     if(rr<=mid)return qur(ls,ll,rr);
 35     if(ll>mid)return qur(rs,ll,rr);
 36     return qur(ls,ll,rr)+qur(rs,ll,rr);
 37 }
 38 void gai(int x,int l,int r,int ll,int rr,int d)
 39 {
 40     if(l>=ll&&r<=rr)
 41     {
 42         lazy[x]+=d;
 43         a[x]+=d*(r-l+1);
 44         return ;
 45     }
 46     int mid=(l+r)>>1;
 47     push_down(x,l,mid,r);
 48     if(ll<=mid)gai(ls,ll,rr,d);
 49     if(rr>mid)gai(rs,ll,rr,d);
 50     a[x]=a[x*2]+a[x*2+1];
 51 }
 52 int tmp[2][N];
 53 void solve(int L,int R,int l,int r)
 54 {
 55     if(L>R)return ;
 56     if(l==r)
 57     {
 58         for(int i=L;i<=R;i++)if(op[q[i]].sign==2)ans[q[i]]=l;
 59         return ;
 60     }
 61     int mid=(l+r)>>1;
 62     int cnt1=0,cnt2=0;
 63     for(int i=L;i<=R;i++)
 64     {
 65         if(op[q[i]].sign==1)
 66         {
 67             if(op[q[i]].c>mid)
 68             {
 69                 gai(1,1,n,op[q[i]].x,op[q[i]].y,1);
 70                 tmp[1][++cnt2]=q[i];
 71             }            
 72             else tmp[0][++cnt1]=q[i];
 73         }
 74         else
 75         {
 76             int y=qur(1,1,n,op[q[i]].x,op[q[i]].y);
 77             if(y>=op[q[i]].c)
 78             {
 79                 tmp[1][++cnt2]=q[i];
 80             }
 81             else op[q[i]].c-=y,tmp[0][++cnt1]=q[i];
 82         }
 83     }
 84     for(int i=L;i<=R;i++)
 85     {
 86         if(op[q[i]].sign==1)
 87         {
 88             if(op[q[i]].c>mid)gai(1,1,n,op[q[i]].x,op[q[i]].y,-1);        
 89         }
 90     }
 91     int l1=L+cnt1-1;
 92     for(int i=1;i<=cnt1;i++)q[L+i-1]=tmp[0][i];
 93     for(int i=1;i<=cnt2;i++)q[l1+i]=tmp[1][i];
 94     solve(L,l1,l,mid);solve(l1+1,R,mid+1,r);
 95 }
 96 signed main()
 97 {
 98     scanf("%lld%lld",&n,&m);
 99     for(int i=1;i<=m;i++)
100     {
101         scanf("%lld%lld%lld%lld",&op[i].sign,&op[i].x,&op[i].y,&op[i].c);
102         q[i]=i;
103     }
104     solve(1,m,1,n);
105     for(int i=1;i<=m;i++)
106     {
107         if(op[i].sign==2)printf("%lld\n",ans[i]);
108     }
109     return 0;
110 }