题目链接：https://vjudge.net/problem/POJ-3667
题目大意：n个连续的房间m个操作。操作分两种，
第一种以1 x形式给出，找到最左的能连续容下x个人的连续房间，并输出左端点的编号，如果找不到就输出0.
第二种以2 l x的形式给出，表示以l为起点的x个房间都清空。
思路：
采用线段树去维护最大连续区间。具体的讲解可以看：https://www.cnblogs.com/lonely-wind-/p/11667471.html
1 #include <math.h>
2 #include <stdio.h>
3 #include <stdlib.h>
4 #include <iostream>
5 #include <algorithm>
6 #include <string>
7 #include <string.h>
8 #include <vector>
9 #include <map>
10 #include <stack>
11 #include <set>
12 //#include <random>
13
14 #define ll long long
15 #define ls nod<<1
16 #define rs (nod<<1)+1
17 const int maxn = 1e5 + 10;
18
19
20 struct segment_tree {
21 int l,r;
22 int lsum,rsum,sum;
23 int lazy;
24 }tree[maxn<<2];
25
26 void pushup(int nod) {
27 int l = tree[nod].l,r = tree[nod].r;
28 int mid = (l + r) >> 1;
29 tree[nod].sum = std::max(std::max(tree[ls].sum,tree[rs].sum),tree[ls].rsum + tree[rs].lsum);
30 tree[nod].lsum = tree[ls].lsum;
31 tree[nod].rsum = tree[rs].rsum;
32 if (tree[ls].lsum == (mid-l+1))
33 tree[nod].lsum = tree[ls].sum + tree[rs].lsum;
34 if (tree[rs].rsum == (r-mid))
35 tree[nod].rsum = tree[rs].sum + tree[ls].rsum;
36 }
37
38 void pushdown(int nod) {
39 int l = tree[nod].l,r = tree[nod].r;
40 int mid = (l + r ) >> 1;
41 tree[ls].sum = tree[ls].rsum = tree[ls].lsum = tree[nod].lazy * (mid-l+1);
42 tree[rs].sum = tree[rs].rsum = tree[rs].lsum = tree[nod].lazy * (r-mid);
43 tree[ls].lazy = tree[rs].lazy = tree[nod].lazy;
44 tree[nod].lazy = -1;
45 }
46
47 void build(int l,int r,int nod) {
48 tree[nod].l = l;
49 tree[nod].r = r;
50 tree[nod].lazy = -1;
51 tree[nod].lsum = tree[nod].rsum = tree[nod].sum = (r - l + 1);
52 if (l == r)
53 return ;
54 int mid = (l + r ) >> 1;
55 build(l,mid,ls);
56 build(mid+1,r,rs);
57 }
58
59 void modify(int x,int y,int z,int nod=1) {
60 int l = tree[nod].l,r = tree[nod].r;
61 if (x <= l && y >= r) {
62 tree[nod].lsum = tree[nod].rsum = tree[nod].sum = (r-l+1)*z;
63 tree[nod].lazy = z;
64 return ;
65 }
66 if (tree[nod].lazy != -1) {
67 pushdown(nod);
68 }
69 int mid = (l + r) >> 1;
70 if (x <= mid) {
71 modify(x,y,z,ls);
72 }
73 if (y > mid) {
74 modify(x,y,z,rs);
75 }
76 pushup(nod);
77 }
78
79 int query(int siz,int nod=1) {
80 int l = tree[nod].l,r = tree[nod].r;
81 int mid = (l + r) >> 1;
82 if (l == r && siz == 1)
83 return l;
84 if (tree[nod].lazy != -1)
85 pushdown(nod);
86 if (tree[nod].sum >= siz) {
87 if (tree[ls].sum >= siz)
88 return query(siz,ls);
89 else if (tree[ls].rsum + tree[rs].lsum >= siz) {
90 return mid + 1 - tree[ls].rsum;
91 }
92 else
93 return query(siz,rs);
94 }
95 return 0;
96 }
97
98 int main() {
99 int n,m;
100 scanf("%d%d",&n,&m);
101 build(1,n,1);
102 while (m--) {
103 int opt;
104 scanf("%d",&opt);
105 if (opt == 1) {
106 int x;
107 scanf("%d",&x);
108 int temp = query(x,1);
109 printf("%d\n",temp);
110 if (!temp) {
111 continue;
112 }
113 modify(temp,temp+x-1,0);
114 }
115 else {
116 int x,y;
117 scanf("%d%d",&x,&y);
118 modify(x,x+y-1,1);
119 }
120 }
121 return 0;
122 }