链接：
http://acm.hdu.edu.cn/showproblem.php?pid=5443
The Water Problem
Time Limit: 1500/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 738 Accepted Submission(s): 591
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
#include<iostream> #include<cstdio> #include<cstring> #include<cstdlib> #include<queue> #include<vector> #include<cmath> using namespace std; #define N 1110 #define INF 0xffffff #define Lson r<<1 #define Rson r<<1|1 #define Mid a[r].mid() struct node { int L, R, Max; int mid() {return (L+R)/2;} }a[N*4]; int n, m; void BuildTree(int r, int L, int R) { a[r].L = L, a[r].R = R, a[r].Max = 0; if(L==R) { scanf("%d", &a[r].Max); return ; } BuildTree(Lson, L, Mid); BuildTree(Rson, Mid+1, R); a[r].Max = max(a[Lson].Max, a[Rson].Max); } int Query(int r, int L, int R) { if(a[r].L==L && a[r].R==R) { return a[r].Max; } if(R<=Mid) return Query(Lson, L, R); else if(L>Mid) return Query(Rson, L, R); else { int LMax = Query(Lson, L, Mid); int RMax = Query(Rson, Mid+1, R); return max(LMax, RMax); } } int main() { int t; scanf("%d", &t); while(t--) { int i, L, R; scanf("%d", &n); BuildTree(1, 1, n); scanf("%d", &m); for(i=1; i<=m; i++) { scanf("%d%d", &L, &R); printf("%d ", Query(1, L, R)); } } return 0; }