题目:
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 22 integers ll and rr, please find out the biggest water source between alal and arar.
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)n(0≤n≤1000) on a line representing the number of water sources. nn 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 qq lines with two integers ll 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
题目大意:
单点修改,区间查询。
解题思路:
RMQ模板题。
实现代码:
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<vector>
#include<cmath>
#include<cstring>
using namespace std;
const int N=1000000+5,LogN=20; //定义整形常量
int logg[N],f[N][LogN+5],a[N];
int n,m,x,y;
int t;
int main(){
scanf("%d",&t);
while(t--){
scanf("%d",&n);
for(int i=1;i<=n;++i){
scanf("%d",&a[i]); //读入n个数据
}
logg[0]=-1; //log[0]=-1,这样才能使log[1]为0;
for(int i=1;i<=n;++i){
f[i][0]=a[i];
logg[i]=logg[i>>1]+1; //预处理长度为1-n的log值
}
for(int j=1;j<=LogN;++j) //计算f[i][j]
for(int i=1;i+(1<<j)-1<=n;++i){ //注意边界不能超过n;
f[i][j]=max(f[i][j-1],f[i+(1<<j-1)][j-1]);
}
scanf("%d",&m);
while(m--){
scanf("%d%d",&x,&y);
int s=logg[y-x+1];
printf("%d\n",max(f[x][s],f[y-(1<<s)+1][s]));
}
}
return 0;
}