zoukankan      html  css  js  c++  java
  • tyvj:1038 忠诚 线段树 区间查询

    题目链接:

    http://www.tyvj.cn/p/1038#

    题意:

    题解:

    区间查询最小值

    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 int a[maxn],ans[maxn];
    19 struct node{
    20     int l,r,minn;
    21 }tree[maxn<<2];
    22 
    23 void pushup(int rt){
    24     tree[rt].minn = min(tree[rt<<1].minn,tree[rt<<1|1].minn);
    25 }
    26 
    27 void build(int rt,int l,int r){
    28     tree[rt].l = l, tree[rt].r = r;
    29     // tree[rt].minn = INF;
    30     if(l == r)
    31         tree[rt].minn = a[l];
    32     else{
    33         int mid = (l+r)/2;
    34         build(rt<<1,l,mid);
    35         build(rt<<1|1,mid+1,r);
    36         pushup(rt);
    37     }
    38 }
    39 
    40 int query(int rt,int l,int r){
    41     int L = tree[rt].l, R = tree[rt].r;
    42     if(l<=L && R<=r)
    43         return tree[rt].minn;
    44     int ans = INF;
    45     int mid = (L+R)/2;
    46     if(l <= mid) ans = min(ans,query(rt<<1,l,r));
    47     if(r>mid) ans = min(ans,query(rt<<1|1,l,r));
    48     return ans;
    49 }
    50 
    51 int main(){
    52     int n=read(), q=read();
    53     for(int i=1; i<=n; i++)
    54         a[i] = read();
    55     build(1,1,n);
    56     int cnt = 0;
    57     for(int i=0; i<q; i++){
    58         int a=read(),b=read();
    59         int tmp = query(1,a,b);
    60         ans[++cnt] = tmp;
    61     }
    62     for(int i=1; i<cnt; i++)
    63         cout << ans[i] << " ";
    64     cout << ans[cnt] << endl;
    65 
    66     return 0;
    67 }
  • 相关阅读:
    Angularjs中的缓存以及缓存清理
    举例子来说明Python引用和对象
    对象关系映射ORM
    Apache Storm 核心概念
    Linux如何查看哪个进程占用的SWAP分区比较多?
    MySQL彻底清除slave信息
    监控MySQL的时候监控用户应该怎么授权?
    MySQL用户密码修改
    专职DBA-Zabbix 3.0 for percona-server TokuDB
    防止rm强制删除
  • 原文地址:https://www.cnblogs.com/yxg123123/p/6827659.html
Copyright © 2011-2022 走看看