POJ 3667 Hotel（成段更新，区间合并）

题意：有N个房间，M次操作。1 a表示找到连续的长度为a的空房间，如果有多解，优先左边的，即表示入住。2 b len把起点为b长度的len的房间清空，即退房。

用co作为延迟标记，如果为-1表示当前区间是两个状态，如果是0表示一种状态，即区间内都没有人入住，如果是1也表示一种状态，表示满客。在线段树节点里保存三个值，端点左边的(lmx)、右边的(rmx)最长的连续的空闲的房间数，总的空闲的房间数目（mx）。查询的时候，依次查询端点左边的连续的空闲房间，左子区间的，左子区间和右子区间连接起来，右区间部分是否满足题意。区间合并还是一样。如果左子区间的lmax等于左子区间的dis，那么父区间的lmx，除了继承左子区间lmx之外，还要加上右子区间的lmx。右端点的同理。

/*代码风格更新后*/#include <iostream>#include <cstdio>#include <cstring>using namespace std;#define LL(x) (x<<1)#define RR(x) (x<<1|1)#define MID(a,b) (a+((b-a)>>1))const int N=50005;struct node{int lft,rht;int flag,mx,lmx,rmx;int len(){return rht-lft+1;}int mid(){return MID(lft,rht);}void init(){mx=lmx=rmx=len();}void fun(int tmp){if(tmp==1) mx=lmx=rmx=0;else mx=lmx=rmx=len();flag=tmp;//忘记添加延迟标记}};int n,m;struct Segtree{node tree[N*4];void down(int ind){if(tree[ind].flag){tree[LL(ind)].fun(tree[ind].flag);tree[RR(ind)].fun(tree[ind].flag);tree[ind].flag=0;}}void up(int ind){tree[ind].lmx=tree[LL(ind)].lmx;tree[ind].rmx=tree[RR(ind)].rmx;tree[ind].mx=max(tree[LL(ind)].mx,tree[RR(ind)].mx);if(tree[LL(ind)].lmx==tree[LL(ind)].len())tree[ind].lmx+=tree[RR(ind)].lmx;if(tree[RR(ind)].rmx==tree[RR(ind)].len())tree[ind].rmx+=tree[LL(ind)].rmx;tree[ind].mx=max(tree[ind].mx,max(tree[ind].lmx,tree[ind].rmx));tree[ind].mx=max(tree[ind].mx,tree[LL(ind)].rmx+tree[RR(ind)].lmx);}void build(int lft,int rht,int ind){tree[ind].lft=lft;tree[ind].rht=rht;tree[ind].flag=0;tree[ind].init();if(lft!=rht){int mid=tree[ind].mid();build(lft,mid,LL(ind));build(mid+1,rht,RR(ind));}}void updata(int st,int ed,int ind,int valu){int lft=tree[ind].lft,rht=tree[ind].rht;if(st<=lft&&rht<=ed) tree[ind].fun(valu);else{down(ind);int mid=tree[ind].mid();if(st<=mid) updata(st,ed,LL(ind),valu);if(ed> mid) updata(st,ed,RR(ind),valu);up(ind);}}int query(int valu,int ind){if(tree[ind].lft==tree[ind].rht) return tree[ind].lft;else{    down(ind);//查询的时候,记得....            int pos;            if(tree[LL(ind)].mx>=valu) pos=query(valu,LL(ind));            else if(tree[LL(ind)].rmx+tree[RR(ind)].lmx>=valu) pos=tree[LL(ind)].rht-tree[LL(ind)].rmx+1;            else pos=query(valu,RR(ind));            up(ind);//查询的时候,记得....            return pos;}}}seg;int main(){while(scanf("%d%d",&n,&m)!=EOF){seg.build(1,n,1);while(m--){int a,b,c,pos=0;scanf("%d",&a);if(a==1){scanf("%d",&b);if(seg.tree[1].mx>=b){pos=seg.query(b,1);seg.updata(pos,pos+b-1,1,1);}printf("%d\n",pos);}else{scanf("%d%d",&b,&c);seg.updata(b,b+c-1,1,-1);}}}return 0;}

/*代码风格更新前*/#include <iostream>#include <cstdio>using namespace std;const int N=50005;struct node{    int left,right,co;    int lmax,rmax,mmax;    int mid(){return left+(right-left)/2;}    int dis(){return right-left+1;}    void change(int a)    {        co=a;        if(co==0) lmax=rmax=mmax=dis();        else lmax=rmax=mmax=0;    }};void unin(node &a,node &b,node &c){    a.lmax=b.lmax;  a.rmax=c.rmax;    a.mmax=max(max(b.mmax,c.mmax),b.rmax+c.lmax);    if(b.lmax==b.dis()) a.lmax+=c.lmax;    if(c.rmax==c.dis()) a.rmax+=b.rmax;}struct Segtree{    node tree[N*4];    void build(int left,int right,int r)    {        tree[r].left=left;  tree[r].right=right;        tree[r].lmax=tree[r].rmax=tree[r].mmax=tree[r].dis();        tree[r].co=-1;        if(left<right)        {            int mid=tree[r].mid();            build(left,mid,r*2);            build(mid+1,right,r*2+1);        }    }    void updata(int be,int end,int r,int co)    {        if(be<=tree[r].left&&tree[r].right<=end)        {            tree[r].change(co);        }        else        {            if(tree[r].co!=-1)            {                tree[r*2].change(tree[r].co);                tree[r*2+1].change(tree[r].co);                tree[r].co=-1;            }            int mid=tree[r].mid();            if(be<=mid) updata(be,end,r*2,co);            if(end>mid) updata(be,end,r*2+1,co);            unin(tree[r],tree[r*2],tree[r*2+1]);        }    }    int query(int len,int r)    {        if(tree[r].lmax>=len) return tree[r].left;        else if(tree[r*2].mmax>=len) return query(len,r*2);        else if(tree[r*2].rmax+tree[r*2+1].lmax>=len) return tree[r*2].right-tree[r*2].rmax+1;        else return query(len,r*2+1);    }}seg;int main(){    int n,m;    scanf("%d%d",&n,&m);    seg.build(1,n,1);    for(int i=0;i<m;i++)    {        int a,b,c;        scanf("%d",&a);        if(a==1)        {            scanf("%d",&b);            if(seg.tree[1].mmax>=b)            {                int pos=seg.query(b,1);                printf("%d\n",pos);                seg.updata(pos,pos+b-1,1,1);            }            else puts("0");        }        else        {            scanf("%d%d",&b,&c);            seg.updata(b,b+c-1,1,0);        }    }    return 0;}