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Time Limit: 1 Sec
Memory Limit: 256 MB
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.
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.
For each query, output an integer representing the size of the biggest water source.
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
题意
查询区间最大值，不带修改
题解：
随便怎么写，反正数据范围只有1000
代码：
#include <cstdio> #include <cmath> #include <cstring> #include <ctime> #include <iostream> #include <algorithm> #include <set> #include <vector> #include <sstream> #include <queue> #include <typeinfo> #include <fstream> #include <map> #include <stack> typedef long long ll; using namespace std; //freopen("D.in","r",stdin); //freopen("D.out","w",stdout); #define sspeed ios_base::sync_with_stdio(0);cin.tie(0) #define test freopen("test.txt","r",stdin) #define maxn 10505 #define mod 10007 #define eps 1e-9 const int inf=0x3f3f3f3f; const ll infll = 0x3f3f3f3f3f3f3f3fLL; inline ll read() { ll x=0,f=1;char ch=getchar(); while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();} while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();} return x*f; } //************************************************************************************** struct node { int l,r; int ma; }; node a[maxn*4]; int num[maxn]; void build(int x,int l,int r) { a[x].l=l,a[x].r=r; if(l==r) { a[x].ma=num[l]; return; } int mid=(l+r)>>1; build(x<<1,l,mid); build(x<<1|1,mid+1,r); a[x].ma=max(a[x<<1].ma,a[x<<1|1].ma); } int query(int x,int l,int r) { int L=a[x].l,R=a[x].r; if(l<=L&&R<=r) return a[x].ma; int mid=(a[x].l+a[x].r)>>1; if(r<=mid) return query(x<<1,l,r); if(l>mid) return query(x<<1|1,l,r); return max(query(x<<1,l,mid),query(x<<1|1,mid+1,r)); } int main() { int t=read(); while(t--) { memset(a,0,sizeof(a)); int n=read(); for(int i=1;i<=n;i++) num[i]=read(); build(1,1,n); int q=read(); for(int i=0;i<q;i++) { int l=read(),r=read(); printf("%d\n",query(1,l,r)); } } }
hdu 5443 The Water Problem 线段树
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原文地址：http://www.cnblogs.com/qscqesze/p/4805295.html
