题目大意：还是一个旅馆里有N个房间，开始均为空。有两种操作：

（1）查询是否有d个连续空房间，有则输出序号最小的房间号，即最靠左的房间号。否则输出0

（2）清空从a房间开始的连续b个房间。

代码如下：

View Code 
  1 #include <stdio.h>
  2 
  3 #define lson l,m,rt<<1
  4 
  5 #define rson m+1,r,rt<<1|1
  6 
  7 #define maxn 50000
  8 
  9 struct node
 10 
 11 {
 12 
 13     int llen,rlen,maxlen,c;
 14 
 15 }setree[maxn<<2];
 16 
 17 int max(int a,int b)
 18 
 19 {
 20 
 21     return a>b?a:b;
 22 
 23 }
 24 
 25 void build(int l,int r,int rt)
 26 
 27 {
 28 
 29     setree[rt].llen=setree[rt].rlen=setree[rt].maxlen=r-l+1;
 30 
 31     setree[rt].c=-1;
 32 
 33     if(l==r)
 34 
 35     return;
 36 
 37     int m=(l+r)>>1;
 38 
 39     build(lson);
 40 
 41     build(rson);
 42 
 43 }
 44 
 45 void pushdown(int rt,int m)
 46 
 47 {
 48 
 49     if(setree[rt].c!=-1){
 50 
 51         if(setree[rt].c==1){
 52 
 53             setree[rt<<1].llen=setree[rt<<1].rlen=setree[rt<<1].maxlen=0;
 54 
 55             setree[rt<<1|1].llen=setree[rt<<1|1].rlen=setree[rt<<1|1].maxlen=0;
 56 
 57             setree[rt<<1].c=setree[rt<<1|1].c=1;
 58 
 59         }
 60 
 61         else if(setree[rt].c==0){
 62 
 63             setree[rt<<1].llen=setree[rt<<1].rlen=setree[rt<<1].maxlen=m-m/2;
 64 
 65             setree[rt<<1|1].llen=setree[rt<<1|1].rlen=setree[rt<<1|1].maxlen=m/2;
 66 
 67             setree[rt<<1].c=setree[rt<<1|1].c=0;
 68 
 69         }
 70 
 71         setree[rt].c=-1;
 72 
 73     }
 74 
 75 }
 76 
 77 void pushup(int rt,int m)
 78 
 79 {
 80 
 81     setree[rt].llen=setree[rt<<1].llen;
 82 
 83     if(setree[rt<<1].llen==m-m/2)
 84 
 85     setree[rt].llen+=setree[rt<<1|1].llen;
 86 
 87     setree[rt].rlen=setree[rt<<1|1].rlen;
 88 
 89     if(setree[rt<<1|1].rlen==m/2)
 90 
 91     setree[rt].rlen+=setree[rt<<1].rlen;
 92 
 93     setree[rt].maxlen=max(setree[rt<<1].rlen+setree[rt<<1|1].llen,max(setree[rt<<1].maxlen,setree[rt<<1|1].maxlen));
 94 
 95 }
 96 
 97 void update(int l,int r,int rt,int L,int R,int c)
 98 
 99 {
100 
101     if(L<=l&&r<=R){
102 
103         setree[rt].llen=setree[rt].rlen=setree[rt].maxlen=c?0:r-l+1;
104 
105         setree[rt].c=c?1:0;
106 
107         return;
108 
109     }
110 
111     pushdown(rt,r-l+1);
112 
113     int m=(l+r)>>1;
114 
115     if(L<=m)
116 
117     update(lson,L,R,c);
118 
119     if(R>m)
120 
121     update(rson,L,R,c);
122 
123     pushup(rt,r-l+1);
124 
125 }
126 
127 int query(int l,int r,int rt,int num)
128 
129 {
130 
131     if(l==r)
132 
133     return l;
134 
135     pushdown(rt,r-l+1);
136 
137     int m=(l+r)>>1;
138 
139     if(setree[rt<<1].maxlen>=num)
140 
141     return query(lson,num);
142 
143     if(setree[rt<<1].rlen+setree[rt<<1|1].llen>=num)
144 
145     return m-setree[rt<<1].rlen+1;
146 
147     return query(rson,num);
148 
149 }
150 
151 int main()
152 
153 {
154 
155     int n,m;
156 
157     scanf("%d%d",&n,&m);
158 
159     build(1,n,1);
160 
161     while(m--){
162 
163         int op,a,b;
164 
165         scanf("%d",&op);
166 
167         if(op==1){
168 
169             scanf("%d",&a);
170 
171             if(a>setree[1].maxlen){
172 
173             printf("0\n");
174 
175             continue;
176 
177             }
178 
179             int temp=query(1,n,1,a);
180 
181             printf("%d\n",temp);
182 
183             update(1,n,1,temp,temp+a-1,1);
184 
185         }
186 
187         else if(op==2){
188 
189             scanf("%d%d",&a,&b);
190 
191             update(1,n,1,a,a+b-1,0);
192 
193         }
194 
195     }
196 
197 }