The Water Problem
Time Limit: 1500/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 2249 Accepted Submission(s): 1757
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 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
Source
Recommend
hujie
區間最大值 RMQ算法預處理即可
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int n;
int dp[50010][50];
void rmq()
{
for(int j=1; (1<<j)<=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 rmq1(int l,int r)
{
int k=0;
while((1<<(k+1))<=r-l+1) k++;
return max(dp[l][k],dp[r-(1<<k)+1][k]);
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
for(int j=1; j<=n; j++)
{
scanf("%d",&dp[j][0]);
}
rmq();
int q;
scanf("%d",&q);
while(q--)
{
int a,b;
scanf("%d%d",&a,&b);
printf("%d\n",rmq1(a,b));
}
}
return 0;
}
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int n;
int dp[50010][50];
void rmq()
{
for(int j=1; (1<<j)<=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 rmq1(int l,int r)
{
int k=0;
while((1<<(k+1))<=r-l+1) k++;
return max(dp[l][k],dp[r-(1<<k)+1][k]);
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
for(int j=1; j<=n; j++)
{
scanf("%d",&dp[j][0]);
}
rmq();
int q;
scanf("%d",&q);
while(q--)
{
int a,b;
scanf("%d%d",&a,&b);
printf("%d\n",rmq1(a,b));
}
}
return 0;
}
人一我百!人十我萬!永不放棄~~~懷着自信的心,去追逐夢想。