//POJ 3667 Hotel 线段树 区间合并(成段更新)/*题目大意：有ｎ间房子，有两种操作:1 a:询问是不是有连续长度为a的空房间,有的话住进最左边2 a b:将[a,a+b-1]的房间清空思路:记录区间中最长的空房间 ，线段树操作:update:区间替换 query:询问满足条件的最左端点 ，线段树每个节点记录 3个参数 ：lsum ：左端开始的空余量 ；rsum ：右端开始的空余量 ；msum : 最大空余量（可能在中间）*/#include<stdio.h>#define N 50005#define lson rt<<1,l,mid#define rson rt<<1|1,mid+1,rint lsum[N<<2],rsum[N<<2],msum[N<<2];int add[N<<2];int n,m;int Max(int x,int y){ return x>y?x:y;}void pushup(int rt,int m){ lsum[rt] = lsum[rt<<1]; rsum[rt] = rsum[rt<<1|1]; if(lsum[rt] == m-(m>>1)) lsum[rt] += lsum[rt<<1|1]; if(rsum[rt] == (m>>1)) rsum[rt] += rsum[rt<<1]; msum[rt] = Max(rsum[rt<<1] + lsum[rt<<1|1],Max(msum[rt<<1],msum[rt<<1|1]));}void pushdown(int rt,int m){ if(add[rt] != -1){ add[rt<<1] = add[rt<<1|1] = add[rt]; lsum[rt<<1] = rsum[rt<<1] = msum[rt<<1] = add[rt] ? 0 : m-(m>>1); lsum[rt<<1|1] = rsum[rt<<1|1] = msum[rt<<1|1] = add[rt] ? 0 : (m>>1); add[rt] = -1; }}void build(int rt,int l,int r){ lsum[rt] = rsum[rt] = msum[rt] = r-l+1; add[rt] = -1; if(l == r) return; int mid = (l + r) >> 1; build(lson); build(rson);}int query(int rt,int l,int r,int w){ if(l == r) return l; pushdown(rt,r-l+1); int mid = (l + r) >> 1; if(msum[rt<<1] >= w) return query(lson,w); else if(rsum[rt<<1] + lsum[rt<<1|1] >= w) return mid-rsum[rt<<1]+1; else return query(rson,w);}void update(int rt,int l,int r,int L,int R,int w){ if(L <= l && R >= r){ lsum[rt] = rsum[rt] = msum[rt] = w ? 0 :r-l+1; add[rt] = w;; return; } pushdown(rt,r-l+1); int mid = (l + r) >> 1; if(L <= mid) update(lson,L,R,w); if(R > mid ) update(rson,L,R,w); pushup(rt,r-l+1);}int main(){ int i; int op,a,b,c; while(scanf("%d %d",&n,&m)!=EOF){ build(1,1,n); for(i = 1; i <= m; ++i){ scanf("%d",&op); if(op == 1){ scanf("%d",&c); if(msum[1] < c) puts("0"); else{ int res = query(1,1,n,c); printf("%d\n",res); update(1,1,n,res,res+c-1,1); } } else{ scanf("%d %d",&a,&b); update(1,1,n,a,a+b-1,0); } } } return 0;}
