ACM-ICPC国际大学生程序设计竞赛北京赛区(2017)网络赛 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

线段树模板题，贴个模板代码。

// Asimple
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstdlib>
#include <queue>
#include <vector>
#include <string>
#include <cstring>
#include <stack>
#include <set>
#include <map>
#include <cmath>
#define INF 0x3f3f3f3f
#define debug(a) cout<<#a<<" = "<<a<<endl
#define test() cout<<"============"<<endl
#define CLS(a,v) memset(a, v, sizeof(a))
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
int dx[] = {-,,,,-,-,,}, dy[]={,,-,,-,,,-};
const int maxn = ;
const ll mod = ;
int n, m, T, len, cnt, num, ans, Max, k;
int x, y;
pair<int, int> dat[*maxn];

void init() {
    for (int i = ; i <  * n - ; ++i) {
        dat[i].first = INF;
        dat[i].second = -INF;
    }
}
void update(int k, int x) {
    k += n - ;
    dat[k] = pair<int, int>(x, x);
    while (k > ) {
        k = (k - ) / ;
        dat[k].first = min(dat[ * k + ].first, dat[ * k + ].first);
        dat[k].second = max(dat[ * k + ].second, dat[ * k + ].second);
    }
}

pair<int, int> query(int a, int b, int k, int l, int r) {
    if (a <= l && r <= b) return dat[k];
    if (a > r || b < l) return pair<int, int>(INF, -INF);
    pair<int, int> vl = query(a, b,  * k + , l, (l + r) / );
    pair<int, int> vr = query(a, b,  * k + , (l + r) /  + , r);
    return pair<int, int>(min(vl.first, vr.first), max(vl.second, vr.second));
}

void input(){
    for(scanf("%d", &T); T--; ) {
        scanf("%d", &k);
        n = (int)pow(, k);
        init();
        for(int i=; i<n; i++) {
            scanf("%d", &num);
            update(i, num);
        }
        for(scanf("%d", &m); m--; ) {
            scanf("%d%d%d",&k, &x, &y);
            if( k ==  ) update(x, y);
            else {
                pair<int, int> p = query(x, y, , , n-);
                ll ans = min((ll)p.first*p.first, min((ll)p.second*p.second, (ll)p.first*p.second));
                printf("%lld\n", ans);
            }
        }
    }
}

int main() {
    input();
    return ;
}