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 }

 

posted @ 2017-03-02 14:33  _yxg123  阅读(106)  评论(0)  编辑  收藏  举报