cdoj 1061 秋实大哥与战争 线段树,合并区间,单点更新,单点查询区间长度

题目链接:

http://acm.uestc.edu.cn/#/problem/show/1061

题意:

题解:

线段树记录从左边延展的长度,记录从右边延展的长度,和区间最大长度
每次更新的时候,就向上更新

代码:

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll;
 4 #define MS(a) memset(a,0,sizeof(a))
 5 #define MP make_pair
 6 #define PB push_back
 7 const int INF = 0x3f3f3f3f;
 8 const ll INFLL = 0x3f3f3f3f3f3f3f3fLL;
 9 inline ll read(){
10     ll x=0,f=1;char ch=getchar();
11     while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
12     while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
13     return x*f;
14 }
15 //////////////////////////////////////////////////////////////////////////
16 const int maxn = 1e5+10;
17 
18 struct node{
19     int l,r,v,lm,rm,mm,len;
20 }tree[maxn<<2];
21 
22 void pushup(int rt){
23     tree[rt].lm = tree[rt<<1].lm;
24     tree[rt].rm = tree[rt<<1|1].rm;
25     tree[rt].mm = max(max(tree[rt<<1].mm,tree[rt<<1|1].mm),tree[rt<<1].rm+tree[rt<<1|1].lm);
26     if(tree[rt<<1].rm == tree[rt<<1].len) tree[rt].lm = tree[rt<<1].lm+tree[rt<<1|1].lm;
27     if(tree[rt<<1|1].lm == tree[rt<<1|1].len) tree[rt].rm = tree[rt<<1|1].rm+tree[rt<<1].rm;
28 }
29 
30 void build(int rt,int l,int r){
31     tree[rt].l = l, tree[rt].r = r;
32     tree[rt].lm = tree[rt].rm = tree[rt].mm = tree[rt].len = r-l+1;
33 
34     if(l != r){
35         int mid = (l+r)/2;
36         build(rt<<1,l,mid);
37         build(rt<<1|1,mid+1,r);
38         pushup(rt);
39     }
40 }
41 
42 void update(int rt,int p,int x){
43     int L = tree[rt].l, R = tree[rt].r;
44     if(L == R){
45         tree[rt].lm=tree[rt].rm=tree[rt].mm=x;
46         return ;
47     }
48     int mid = (L+R)/2;
49     if(p<=mid) update(rt<<1,p,x);
50     else update(rt<<1|1,p,x);
51     pushup(rt);
52 }
53 
54 ll query(int rt,int p){
55     int L = tree[rt].l, R = tree[rt].r;
56     if(L==R || tree[rt].mm==0 || tree[rt].mm==tree[rt].len)
57         return tree[rt].mm;
58     int mid = (L+R)/2;
59     if(p <= mid){
60         if(p>=tree[rt<<1].r-tree[rt<<1].rm+1)
61             return query(rt<<1,p) + query(rt<<1|1,mid+1);
62         else
63             return query(rt<<1,p);
64     }else{
65         if(p<=tree[rt<<1|1].l+tree[rt<<1|1].lm-1)
66             return query(rt<<1|1,p) + query(rt<<1,mid);
67         else
68             return query(rt<<1|1,p);
69     }
70 }
71 
72 int main(){
73     int n=read(),m=read();
74     build(1,1,n);
75     for(int i=0; i<m; i++){
76         int op = read(), x = read();
77         if(op == 0){
78             update(1,x,0);
79         }else if(op == 1){
80             update(1,x,1);
81         }else{
82             ll ans = query(1,x);
83             printf("%lld\n",ans);
84         }
85     }
86 
87     return 0;
88 }

 

posted @ 2017-03-02 18:51  _yxg123  阅读(139)  评论(0编辑  收藏  举报