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);
}
}
}
}