Hdoj The Water problem 5443 (segment tree interval max-value problem)

The water problem Time limit:1500/1000 MS (java/others) Memory limit:131072/131072 K (java/others)
Total submission (s): 921 Accepted Submission (s): 731


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 is given an integer T (t≤10) indicating the number of test cases. For each test case, there was 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 are in {1,..., 106}. On the next line, there is a number Q (0≤q≤1000) representing the number of queries. After that, there'll be Q lines with the 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 <stdio.h> #include <string.h> #include <algorithm> #define N 1010 using namespace std;
int a[n];
	struct ZZ {int l;
	int R;
int m;
}q[4*n];
	void build (int gen,int l,int r) {q[gen].l=l;
	Q[gen].r=r;
		if (l==r) {q[gen].m=a[l];
	return;
	} int mid= (L+R)/2;
	Build (Gen<<1,l,mid);
	Build (Gen<<1|1,mid+1,r);
Q[gen].m=max (Q[GEN&LT;&LT;1].M,Q[GEN&LT;&LT;1|1].M);
	} int query (int gen,int l,int R) {if (L&LT;=Q[GEN].L&AMP;&AMP;R&GT;=Q[GEN].R) return q[gen].m;
	int mid= (Q[GEN].L+Q[GEN].R)/2;
	int cnt;
	if (r<=mid) cnt=query (gen<<1,l,r);
	else if (l>=mid+1) cnt=query (gen<<1|1,l,r);
	else Cnt=max (query (gen<<1,l,mid), query (Gen<<1|1,mid+1,r));
return CNT;
	} int main () {int t,n,m,i;
	int x, y;
	scanf ("%d", &t);
		while (t--) {scanf ("%d", &n);
		for (i=1;i<=n;i++) scanf ("%d", &a[i]);
		Build (1,1,n);
		scanf ("%d", &m);
			while (m--) {scanf ("%d%d", &x,&y);
		printf ("%d\n", Query (1,x,y));
	}} return 0; }