hdu 5443 The Water Problem(求区间最值+ST表)

The Water Problem

Time Limit: 1500/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1196    Accepted Submission(s): 942

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

 

没什么说的，ST水题，或者用线段树也可以。

#include<cstdio>
#include<algorithm>
using namespace std;
const int maxn = 1550;
int dp[maxn][20],mm[maxn];
int a[maxn], n;
void init()
{
    mm[0] = -1;
    for (int i = 1; i <= n; i++)
    {
        mm[i] = ((i&(i - 1)) == 0) ? mm[i - 1] + 1 : mm[i - 1];
        dp[i][0] = a[i];
    }
    for (int j = 1; j <= mm[n]; j++)
        for (int i = 1; i + (1 << j) - 1 <= n; i++)
        dp[i][j] = max(dp[i][j - 1], dp[i + (1 << (j - 1))][j - 1]);
}
int rmq(int x, int y)
{
    int k = mm[y - x + 1];
    return max(dp[x][k], dp[y - (1 << k) + 1][k]);
}
int main()
{
    int t,q,x,y;
    scanf("%d", &t);
    while (t--)
    {
        scanf("%d", &n);
        for (int i = 1; i <= n; i++)
            scanf("%d", &a[i]);
        init();
        scanf("%d", &q);
        while (q--)
        {
            scanf("%d%d", &x, &y);
            printf("%d\n", rmq(x, y));
        }
    }
    return 0;
}