poj 3667 Hotel

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线段树之区间合并，有一个线段，从1到n，下面m个操作，操作分两个类型，以1开头的是查询操作，以2开头的是更新操作

1 w 表示在总区间内查询一个长度为w的可用区间，并且要最靠左，能找到的话返回这个区间的左端点并占用了这个区间，找不到返回0

接近2个月没怎么学算法刷题了，也落下了很多东西，慢慢补吧，明天开始二维线段树

#include<iostream>#define maxn 50010using namespace std;int n,m,a,b;int cmd;struct stu{int r,l;int flag;int tlen,llen,rlen;int up(){tlen=llen=rlen=(flag? 0:r-l+1);}};stu mapp[maxn*4];void build(int l,int r,int count){mapp[count].l=l;mapp[count].r=r;mapp[count].tlen=mapp[count].llen=mapp[count].rlen=r-1+1;mapp[count].flag=0;if(l==r) return;int mid=(l+r)/2;build(l,mid,count*2);build(mid+1,r,count*2+1);}void push(int count,int x){if(mapp[count].flag!=-1){mapp[count*2].flag=mapp[count*2+1].flag=mapp[count].flag;mapp[count].flag=-1;mapp[count*2].up();mapp[count*2+1].up();}}int que(int l,int r,int count){//if(mapp[count].l==mapp[count].r&&mapp[count].tlen) return mapp[count].l;push(count,r-l+1);if(mapp[count*2].tlen>=a) return que(l,r,count*2);else if(mapp[count*2].rlen+mapp[count*2+1].llen>=a){return mapp[count*2].r-mapp[count*2].rlen+1;}else if(mapp[count*2+1].tlen>=a) return que(l,r,count*2+1);else return 0;}void updata(int l,int r,int v,int count){if(mapp[count].l==l&&mapp[count].r==r){mapp[count].flag=v;mapp[count].up();return;}push(count,r-l+1);int mid=(mapp[count].l+mapp[count].r)/2;if(r<=mid) updata(l,r,v,count*2);else if(l>=mid+1) updata(l,r,v,count*2+1);else{updata(l,mid,v,count*2);updata(mid+1,r,v,count*2+1);}int tmp=max(mapp[count*2].tlen,mapp[count*2+1].tlen);mapp[count].tlen=max(tmp,mapp[count*2].rlen+mapp[count*2+1].llen);mapp[count].llen=mapp[count*2].llen;mapp[count].rlen=mapp[count*2+1].rlen;if(mapp[count*2].tlen==(mapp[count*2].r-mapp[count*2].l+1)){mapp[count].llen+=mapp[count*2+1].llen;}if(mapp[count*2+1].tlen==(mapp[count*2+1].r-mapp[count*2+1].l+1)){mapp[count].rlen+=mapp[count*2].rlen;}}int main(){cin.sync_with_stdio(false);while(cin>>n>>m){build(1,n,1);for(int i=0;i<m;i++){cin>>cmd;if(cmd==1){cin>>a;int ans=que(1,n,1);cout<<ans<<endl;if(ans) updata(ans,ans+a-1,1,1);}else{cin>>a>>b;updata(a,a+b-1,0,1);}}}return 0;}

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