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,...,anrepresenting
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;
}