ACM-ICPC北京赛区(2017)网络赛_Minimum

Posted 2020-10-08 _季末

题目9 : Minimum

时间限制:1000ms

单点时限:1000ms

内存限制:256MB

描述

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.

输入

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)

输出

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

样例输入

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

样例输出

1
1
4

典型的线段树单点更新与区间最大（小）值，没什么奥妙，但是自己还是打挫了，连WA带T13发，233，最终在一位小姐姐的指导小改掉一些细节错误成功A一发

附上AC代码：

  1 #include<iostream>
  2 #include<cstdio>
  3 #include<cmath>
  4 #include<string>
  5 #include<cstring>
  6 
  7 using namespace std;
  8 
  9 const int MAXNODE = (1<<20)+5;
 10 const int MAX = 2e7+10;//2*10^6+10
 11 const int INF = 0x7fffffff;
 12 
 13 struct NODE{
 14     int maxvalue, minvalue;
 15     int left, right;
 16 }node[MAXNODE];
 17 
 18 int father[MAX];
 19 
 20 
 21 void BuildTree(int i, int left, int right)
 22 {
 23     node[i].minvalue = INF;
 24     node[i].maxvalue = -INF;
 25     node[i].left = left;
 26     node[i].right = right;
 27     if(right == left)
 28     {
 29         father[left] = i;
 30         return;
 31     }
 32     BuildTree(i<<1, left, (int)(floor(left+right)/2.0));
 33     BuildTree((i<<1)+1, (int)(floor(left+right)/2.0)+1, right);
 34 }
 35 
 36 
 37 //自底向上更新
 38 void UpdateTree(int ri)
 39 {
 40     if(ri <= 1)
 41         return;
 42     int fi = ri/2;
 43     int a = node[fi<<1].maxvalue;
 44     int b = node[(fi<<1)+1].maxvalue;
 45     node[fi].maxvalue = max(a, b);
 46     a = node[fi<<1].minvalue;
 47     b = node[(fi<<1)+1].minvalue;
 48     node[fi].minvalue = min(a, b);
 49     UpdateTree(ri/2);
 50 }
 51 
 52 
 53 int Max, Min;
 54 
 55 
 56 //自顶向上查询
 57 void Query_max(int i, int l, int r)
 58 {
 59     if(node[i].left == l && node[i].right == r)
 60     {
 61         Max = max(Max, node[i].maxvalue);
 62         return;
 63     }
 64 
 65     i <<= 1;
 66     if(l <= node[i].right)
 67     {
 68         if(r <= node[i].right)
 69         {
 70             Query_max(i, l, r);
 71         }
 72         else
 73         {
 74             Query_max(i, l, node[i].right);
 75         }
 76     }
 77     i++;
 78     if(r >= node[i].left)
 79     {
 80         if(l >= node[i].left)
 81         {
 82             Query_max(i, l, r);
 83         }
 84         else
 85         {
 86             Query_max(i, node[i].left, r);
 87         }
 88     }
 89 }
 90 
 91 
 92 void Query_min(int i, int l, int r)
 93 {
 94     if(node[i].left == l && node[i].right == r)
 95     {
 96         Min = min(Min, node[i].minvalue);
 97         return;
 98     }
 99 
100     i <<= 1;
101     if(l <= node[i].right)
102     {
103         if(r <= node[i].right)
104         {
105             Query_min(i, l, r);
106         }
107         else
108         {
109             Query_min(i, l, node[i].right);
110         }
111     }
112     i++;
113     if(r >= node[i].left)
114     {
115         if(l >= node[i].left)
116         {
117             Query_min(i, l, r);
118         }
119         else
120         {
121             Query_min(i, node[i].left, r);
122         }
123     }
124 }
125 
126 
127 int main()
128 {
129     int k, t, n, g, a, b, c;
130     long long ans;
131     ios::sync_with_stdio(false);
132     cin>>t;
133     //cout<<INF<<-INF<<endl;
134     while(t--)
135     {
136         cin>>k;
137         BuildTree(1, 1, 1<<k);
138         for(int i = 1; i <= 1<<k; i++)
139         {
140             cin>>g;
141             node[father[i]].maxvalue = node[father[i]].minvalue = g;
142             UpdateTree(father[i]);
143         }
144         cin>>n;
145         for(int i = 1; i <= n; i++)
146         {
147             Max = -INF;
148             Min = INF;
149             cin>>a>>b>>c;
150             if(a == 1)
151             {
152                 Query_max(1, b+1, c+1);
153                 Query_min(1, b+1, c+1);
154                 //cout<<Max<<" "<<Min<<endl;
155                 if(Max <= 0)
156                 {
157                     ans = (long long)Max*Max;
158                     cout<<ans<<endl;
159                 }
160                 else if(Min >= 0)
161                 {
162                     ans = (long long)Min*Min;
163                     cout<<ans<<endl;
164                 }
165                 else
166                 {
167                     ans = (long long)Min*Max;
168                     cout<<ans<<endl;
169                 }
170             }
171             else if(a == 2)
172             {
173                 node[father[b+1]].maxvalue = node[father[b+1]].minvalue = c;
174                 UpdateTree(father[b+1]);
175             }
176         }
177     }
178 return 0;
179 }