2015长春网络赛 1007 - The Water Problem(裸线段树)

The Water Problem

Time Limit: 1500/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 0    Accepted Submission(s): 0

Problem Description

In Land waterless, water is a very limited resource. People always fight for the biggest source of water. Given a sequence of water sources with  a1,a2,a3,...,an representing the size of the water source. Given a set of queries each containing  2  integers  l  and  r , please find out the biggest water source between  al  and  ar .

 

Input

First you are given an integer  T(T≤10)  indicating the number of test cases. For each test case, there is a number  n(0≤n≤1000)  on a line representing the number of water sources.  n  integers follow, respectively  a1,a2,a3,...,an , and each integer is in  {1,...,106} . On the next line, there is a number  q(0≤q≤1000)  representing the number of queries. After that, there will be  q  lines with two integers  l  and  r(1≤l≤r≤n)  indicating the range of which you should find out the biggest water source.

 

Output

For each query, output an integer representing the size of the biggest water source.

 

Sample Input

   
   
   
   

    
    
    
    3
1
100
1
1 1
5
1 2 3 4 5
5
1 2
1 3
2 4
3 4
3 5
3
1 999999 1
4
1 1
1 2
2 3
3 3
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    100
2
3
4
4
5
1
999999
999999
1
   
   
   
   
   
   
   
   

    
    
    
    由于题目一样，水题
   
   
   
   
   
   
   
   

    
    
    
    
#include <map>
#include <set>
#include <cmath>
#include <ctime>
#include <queue>
#include <vector>
#include <cctype>
#include <cstdio>
#include <string>
#include <cstring>
#include <sstream>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;


#define pb push_back
#define mp make_pair
#define fillchar(a, x) memset(a, x, sizeof(a))
#define copy(a, b) memcpy(a, b, sizeof(a))
#define lson rt << 1, l, mid
#define rson rt << 1|1, mid + 1, r


typedef long long LL;
typedef pair<int, int > PII;
typedef unsigned long long uLL;
template<typename T>
void print(T* p, T* q, string Gap = " ") {
    int d = p < q ? 1 : -1;
    while(p != q) {
        cout << *p;
        p += d;
        if(p != q) cout << Gap;
    }
    cout << endl;
}
template<typename T>
void print(const T &a, string bes = "") {
    int len = bes.length();
    if(len >= 2)cout << bes[0] << a << bes[1] << endl;
    else cout << a << endl;
}

const int INF = 0x3f3f3f3f;
const int MAXM = 2e5;
const int MAXN = 1e3 + 5;

int T, n , Sum[MAXN << 2];

void pushup(int rt) {
    Sum[rt] = max(Sum[rt << 1], Sum[rt << 1|1]);
}

void build(int rt, int l, int r) {
    if(l == r){
        cin >> Sum[rt];
        return;
    }
    int mid = (l + r) >> 1;
    build(lson);
    build(rson);
    pushup(rt);
}

int query(int L, int R, int rt, int l, int r) {
    if(L <= l && r <= R) {
        return Sum[rt];
    }
    int mid = (l + r) >> 1;
    int ret = 0;
    if(L <= mid) ret = max(ret, query(L, R, lson));
    if(R > mid) ret = max(ret, query(L, R, rson));
    return ret;
}

int main(){
    //freopen("D://imput.txt", "r", stdin);
    int l, r, q;
    cin >> T;
    while(T --){
        cin >> n;
        build(1, 1, n);
        cin >> q;
        while(q --){
            cin >> l >> r;
            cout << query(l, r, 1, 1, n) << endl;
        }
    }
    return 0;
}