POJ 3667

  题意：  一个旅馆有N个连续的空房间。。。

  接下来一些操作   

    进来的旅客都需要有一些连续的空房间

如果有则输出首个房间的位置。如果没有就输出0.

  或者是在某个位置开始空出连续的一些房间。（房间原来就可以是空的）  

思路：

      首先建立线段树。。

    每个区间要保存该区间内从左数的空房间数,lk,从右数的空房间数rk和最大空房间数mk.

每次更新改区间的时候就是分情况来更新lk,rk,mk这三个数据。

  剩下的没有什么需要注意的了。。。只要注意查找区间时的顺序应该是从左子树，左子树的rk+右子树的lk，右子树就可以了。

贴上代码。

#include<stdio.h>#include<string.h>#include<iostream>#include<algorithm>using namespace std;int m;struct tree{   int left;   int right;   int lk;   int mk;   int rk;   int flag;   int lazy;}tree[800000];void inset(int inst,int le,int ri){    tree[inst].left=le;    tree[inst].right=ri;    tree[inst].lk=(ri-le+1);    tree[inst].rk=(ri-le+1);    tree[inst].mk=(ri-le+1);    tree[inst].flag=0;    tree[inst].lazy=2;    if(le==ri) return ;    int mid=(le+ri)>>1;    inset(2*inst,le,mid);    inset(2*inst+1,mid+1,ri);}void cd(int inst){  //cout<<"cd"<<endl;    if(tree[inst].lazy==0)    {  //cout<<"cd1 "<<inst<<endl;        tree[2*inst].lk=(tree[2*inst].right-tree[2*inst].left+1);        tree[2*inst].rk=(tree[2*inst].right-tree[2*inst].left+1);        tree[2*inst].mk=(tree[2*inst].right-tree[2*inst].left+1);        tree[2*inst+1].lk=(tree[2*inst+1].right-tree[2*inst+1].left+1);        tree[2*inst+1].rk=(tree[2*inst+1].right-tree[2*inst+1].left+1);        tree[2*inst+1].mk=(tree[2*inst+1].right-tree[2*inst+1].left+1);        tree[inst].flag=1;        tree[2*inst].lazy=0;tree[2*inst+1].lazy=0;        tree[inst].lazy=2;    }    else if(tree[inst].lazy==1)    { //cout<<"cd2 "<<inst<<endl;        tree[2*inst].lk=0;        tree[2*inst].mk=0;        tree[2*inst].rk=0;        tree[2*inst+1].lk=0;        tree[2*inst+1].mk=0;        tree[2*inst+1].rk=0;        tree[inst].flag=1;        tree[2*inst].lazy=1;tree[2*inst+1].lazy=1;        tree[inst].lazy=2;    }}void add(int inst,int le,int ri){    if(tree[inst].left==le&&tree[inst].right==ri)    {        tree[inst].lk=(ri-le+1);        tree[inst].mk=(ri-le+1);        tree[inst].rk=(ri-le+1);        tree[inst].lazy=0;        tree[inst].flag=0;        //cout<<tree[inst].mk<<endl;        return ;    }    cd(inst);    tree[inst].flag=1;    if(tree[inst].left==tree[inst].right) return ;    int mid=(tree[inst].left+tree[inst].right)>>1;    if(ri<=mid)        add(2*inst,le,ri);    else if(le>mid)        add(2*inst+1,le,ri);    else    {        add(2*inst,le,mid);        add(2*inst+1,mid+1,ri);    }}void add1(int inst,int le,int ri){     //cout<<inst<<" "<<le<<" "<<ri<<endl;    if(tree[inst].left==le&&tree[inst].right==ri)    {  //cout<<tree[inst].left<<" "<<tree[inst].right<<endl;        tree[inst].lk=0;        tree[inst].mk=0;        tree[inst].rk=0;        tree[inst].lazy=1;        tree[inst].flag=0;        return ;    }     cd(inst);     tree[inst].flag=1;     if(tree[inst].left==tree[inst].right) return ;    int mid=(tree[inst].left+tree[inst].right)>>1;    if(ri<=mid)        add1(2*inst,le,ri);    else if(le>mid)        add1(2*inst+1,le,ri);    else    {        add1(2*inst,le,mid);        add1(2*inst+1,mid+1,ri);    }}void flag(int inst){  //cout<<"flag"<<endl;    if(tree[inst].lazy!=2) cd(inst);    //if(inst==3) cout<<tree[2*inst].mk<<" "<<tree[2*inst+1].mk<<"  1"<<endl;    if(tree[2*inst].flag!=0)        flag(2*inst);    if(tree[2*inst+1].flag!=0)        flag(2*inst+1);    tree[inst].flag=0;    tree[inst].lazy=2;    tree[inst].lk=tree[2*inst].lk;    tree[inst].rk=tree[2*inst+1].rk;    tree[inst].mk=max(tree[2*inst].mk,max(tree[2*inst+1].mk,tree[2*inst].rk+tree[2*inst+1].lk));    //if(inst==3) cout<<tree[inst].mk<<"  2"<<endl;    if(tree[2*inst].lk==(tree[2*inst].right-tree[2*inst].left+1))        tree[inst].lk=tree[2*inst].lk+tree[2*inst+1].lk;    if(tree[2*inst+1].rk==(tree[2*inst+1].right-tree[2*inst+1].left+1))        tree[inst].rk=tree[2*inst].rk+tree[2*inst+1].rk;        //cout<<inst<<" "<<tree[inst].lk<<" "<<tree[2*inst].lk<<"  lk"<<endl;        //cout<<inst<<" "<<tree[inst].rk<<" "<<tree[2*inst+1].rk<<"  rk"<<endl;        //cout<<inst<<" "<<tree[inst].mk<<" "<<tree[2*inst].mk<<"  mk"<<endl;}void query(int inst,int k){    //cout<<tree[3].mk<<" 1"<<endl;    if(tree[inst].flag!=0)        flag(inst);        //cout<<tree[3].mk<<" 2"<<endl;    //cout<<inst<<" "<<tree[inst].mk<<" "<<tree[2*inst].mk<<" "<<tree[2*inst+1].mk<<endl;    if(tree[inst].right==tree[inst].left)    {  if(k==1)       {   if(tree[inst].mk==1)          {              m=tree[inst].left;              //cout<<"add1"<<tree[inst].left<<" "<<tree[inst].right<<endl;           add1(inst,tree[inst].left,tree[inst].right);          }          else m=0;       }        else m=0;        //cout<<m<<endl;        return ;    }         cd(inst);    if(tree[inst].mk>=k)        {   tree[inst].flag=1;            tree[inst].lazy=2;            if(tree[2*inst].mk>=k)            {  //cout<<m<<endl;                query(2*inst,k);            }            else if(tree[2*inst].rk+tree[2*inst+1].lk>=k&&tree[2*inst].rk!=0)            {     m=tree[2*inst].right-tree[2*inst].rk+1;                    int t=tree[2*inst+1].left+k-tree[2*inst].rk-1;                add1(inst,tree[2*inst].right-tree[2*inst].rk+1,tree[2*inst].right);                add1(inst,tree[2*inst+1].left,t);                //cout<<inst<<endl;            }            else if(tree[2*inst+1].mk>=k)            { //cout<<inst<<endl;                query(2*inst+1,k);            }        }    else {  //cout<<"###"<<endl;            m=0; return ;    }}int main(){    int n,m1;    while(scanf("%d%d",&n,&m1)!=EOF)    {        inset(1,1,n);        while(m1--)        {            int x,a,b;            scanf("%d",&x);            if(x==1)            {   m=0;                scanf("%d",&a);                //cout<<a<<endl;                query(1,a);                printf("%d\n",m);               // cout<<"!!!  "<<tree[3].mk<<endl;            }            else            {                scanf("%d%d",&a,&b);                add(1,a,a+b-1);            }        }    }}