hdu4614 Vases and Flowers 线段树

Alice is so popular that she can receive many flowers everyday. She has N vases numbered from 0 to N-1. When she receive some flowers, she will try to put them in the vases, one flower in one vase. She randomly choose the vase A and try to put a flower in the vase. If the there is no flower in the vase, she will put a flower in it, otherwise she skip this vase. And then she will try put in the vase A+1, A+2, ..., N-1, until there is no flower left or she has tried the vase N-1. The left flowers will be discarded. Of course, sometimes she will clean the vases. Because there are too many vases, she randomly choose to clean the vases numbered from A to B(A <= B). The flowers in the cleaned vases will be discarded.

题意：有 0~n-1 共 n 个花瓶，一个花瓶可以插一束花，现在有两个操作，一个是从 A 号花瓶开始插 k 束花，若花瓶有花就顺延到下一个花瓶，直到插完 k 束花或插完 n-1 号花瓶，问她插的第一个和最后一个花瓶编号。第二个操作是清空一段区间的花，问共丢掉多少花。

线段树保存区间空瓶数、首个空瓶和末个空瓶。树上二分查找 k 个空瓶，并更新最大最小空瓶位置。区间清除。

 #include<stdio.h>
 #include<string.h>
 const int maxm=;
 const int INF=0x3f3f3f3f;

 int st[maxm<<],ma[maxm<<],mi[maxm<<];
 int ch[maxm<<];
 int maxx,minn,k;

 inline int max(int a,int b){return a>b?a:b;}
 inline int min(int a,int b){return a<b?a:b;}

 void build(int o,int l,int r){
     ma[o]=r;
     mi[o]=l;
     st[o]=r-l+;
     ch[o]=;
     if(l==r)return;
     int m=l+((r-l)>>);
     build(o<<,l,m);
     build(o<<|,m+,r);
 }

 void pushdown(int o,int l,int r){
     if(ch[o]==){
         ch[o<<]=ch[o<<|]=;
         int m=l+((r-l)>>);
         st[o<<]=st[o<<|]=;
         ma[o<<]=ma[o<<|]=-;
         mi[o<<]=mi[o<<|]=INF;
         ch[o]=;
     }
     else if(ch[o]==){
         ch[o<<]=ch[o<<|]=;
         int m=l+((r-l)>>);
         st[o<<]=m-l+;
         st[o<<|]=r-m;
         ma[o<<]=m;
         ma[o<<|]=r;
         mi[o<<]=l;
         mi[o<<|]=m+;
         ch[o]=;
     }
 }

 void update1(int o,int l,int r,int ql,int qr){
     if(ql<=l&&qr>=r){
         if(k>=st[o]){
             k-=st[o];
             maxx=max(maxx,ma[o]);
             minn=min(minn,mi[o]);
             ma[o]=-;
             mi[o]=INF;
             st[o]=;
             ch[o]=;
             return;
         }
         else if(l==r)return;
     }
     pushdown(o,l,r);
     int m=l+((r-l)>>);
     if(ql<=m&&k)update1(o<<,l,m,ql,qr);
     if(qr>=m+&&k)update1(o<<|,m+,r,ql,qr);
     ma[o]=max(ma[o<<],ma[o<<|]);
     mi[o]=min(mi[o<<],mi[o<<|]);
     st[o]=st[o<<]+st[o<<|];
 }

 int update2(int o,int l,int r,int ql,int qr){
     if(ql<=l&&qr>=r){
         int ans=r-l+-st[o];
         st[o]=r-l+;
         ma[o]=r;
         mi[o]=l;
         ch[o]=;
         return ans;
     }
     pushdown(o,l,r);
     int m=l+((r-l)>>);
     int ans=;
     if(ql<=m)ans+=update2(o<<,l,m,ql,qr);
     if(qr>=m+)ans+=update2(o<<|,m+,r,ql,qr);
     ma[o]=max(ma[o<<],ma[o<<|]);
     mi[o]=min(mi[o<<],mi[o<<|]);
     st[o]=st[o<<]+st[o<<|];
     return ans;
 }

 int main(){
     int T;
     scanf("%d",&T);
     while(T--){
         int n,m;
         scanf("%d%d",&n,&m);
         build(,,n-);
         for(int i=;i<=m;++i){
             int t;
             scanf("%d",&t);
             if(t==){
                 int a;
                 scanf("%d%d",&a,&k);
                 maxx=-;
                 minn=INF;
                 update1(,,n-,a,n-);
                 if(maxx==-&&minn==INF)printf("Can not put any one.\n");
                 else printf("%d %d\n",minn,maxx);
             }
             else if(t==){
                 int a,b;
                 scanf("%d%d",&a,&b);
                 printf("%d\n",update2(,,n-,a,b));
             }
         }
         printf("\n");
     }
     return ;
 }