hihoCoder #1586 : Minimum-结构体版线段树(单点更新+区间最值求区间两数最小乘积) (ACM-ICPC国际大学生程序设计竞赛北京赛区(2017)网络赛)

#1586 : Minimum

Time Limit:1000ms

Case Time Limit:1000ms

Memory Limit:256MB

Description

You are given a list of integers a0, a1, …, a2^k-1.

You need to support two types of queries:

1. Output Minx,y∈[l,r] {ax∙ay}.

2. Let ax=y.

Input

The first line is an integer T, indicating the number of test cases. (1≤T≤10).

For each test case:

The first line contains an integer k (0 ≤ k ≤ 17).

The following line contains 2k integers, a0, a1, …, a2^k-1 (-2k ≤ ai < 2k).

The next line contains a integer  (1 ≤ Q < 2k), indicating the number of queries. Then next Q lines, each line is one of:

1. 1 l r: Output Minx,y∈[l,r]{ax∙ay}. (0 ≤ l ≤ r < 2k)

2. 2 x y: Let ax=y. (0 ≤ x < 2k, -2k ≤ y < 2k)

Output

For each query 1, output a line contains an integer, indicating the answer.

Sample Input

1
3
1 1 2 2 1 1 2 2
5
1 0 7
1 1 2
2 1 2
2 2 2
1 1 2

Sample Output

1
1
4

 

 

 

这个题就是区间查询最大值和最小值，通过判断区间最大值和最小值的正负来得到区间最小乘积。

代码:

 #include<iostream>
 #include<cstring>
 #include<cstdio>
 #include<cstdlib>
 #include<algorithm>
 #include<cmath>
 #include<queue>
 #include<vector>
 #define inf 0x7fffffff
 #define lson l,m,rt<<1
 typedef long long ll;
 #define rson m+1,r,rt<<1|1
 using namespace std;
 const int maxn=1e6+;
 struct node{
     int left,right,maxx,minn;
 }tree[maxn*];
 int st_min[maxn<<],st_max[maxn<<];
 inline int minn(int a,int b){return a>b?b:a;}
 inline int maxx(int a,int b){return a>b?a:b;}
 void PushUP(int rt){
     st_min[rt]=minn(st_min[rt<<],st_min[rt<<|]);
     st_max[rt]=maxx(st_max[rt<<],st_max[rt<<|]);
 }
 void build(int l,int r,int rt) {
     if(l==r){
         scanf("%d",&st_min[rt]);
         st_max[rt]=st_min[rt];
         return ;
     }
     int m=(l+r)>>;
     build(lson);
     build(rson);
     PushUP(rt);
 }
 void update(int p,int num,int l,int r,int rt){
        if(l==r){
         st_max[rt]=num;
         st_min[rt]=num;
         return ;
        }
        int m=(l+r)>>;
        if(p<=m)update(p,num,lson);
        else update(p,num,rson);
        PushUP(rt);
 }
 int query_min(int L,int R,int l,int r,int rt){
     if (L<=l&&r<=R){
         return st_min[rt];
     }
     int m=(l+r)>>;
     int ret1=inf,ret2=inf;
     if(L<=m)ret1=query_min(L,R,lson);
     if(R>m) ret2=query_min(L,R,rson);
     return minn(ret1,ret2);
 }
 int query_max(int L,int R,int l,int r,int rt){
     if(L<=l&&r<=R){
        return st_max[rt];
     }
     int m=(l+r)>>;
     int ret1=-inf,ret2=-inf;
     if(L<=m)ret1=query_max(L,R,lson);
     if(R>m) ret2=query_max(L,R,rson);
     return maxx(ret1,ret2);
 }
 int main(){
     int t;
     scanf("%d",&t);
     while(t--){
         int n,m;
         scanf("%d",&n);
         int k=pow(,n);
         build(,k,);
         scanf("%d",&m);
         while(m--){
             int h,a,b;
             scanf("%d%d%d",&h,&a,&b);
             if(h==)update(a+,b,,k,);
             else{
                 ll ans1=query_min(a+,b+,,k,);
                 ll ans2=query_max(a+,b+,,k,);
                 if(ans1<&&ans2>=)
                     printf("%lld\n",ans2*ans1);
                 else if(ans1<&&ans2<)
                     printf("%lld\n",ans2*ans2);
                 else
                     printf("%lld\n",ans1*ans1);
             }
         }
     }
     return ;
 }