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
; 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;
}