ST表是一个用来解决 RMQ(区间最值)问题的算法,以求最小值为例:
一个区间可以分为左区间和右区间
区间最小值=min(左区间最小值,右区间最小值)
令dp[i][j] 表示[i,i+2^j-1]区间的最小值,则 dp[i][j]=min(dp[i][j-1],dp[i+(1<<(j-1))][j-1])
void init()
{
//初始化
for(int j=1;j<=21;j++)
for(int i=1;i+(1<<j)-1<=n;i++)
dp[i][j]=min(dp[i][j-1],dp[i+(1<<(j-1))][j-1]);
}
一个长度为n的区间:k=log2(n), dp[i][j]=min(dp[i][k],dp[i+n-(1<<k)][k])
#include <iostream>
#include<cmath>
using namespace std;
int n,m;
int dp[100005][25];
void init()
{
int i,j;
for(j=1;j<=17;j++)
{
for(i=1;i<=n;i++)
{
if(i+(1<<j)-1<=n)
dp[i][j]=min(dp[i][j-1],dp[i+(1<<(j-1))][j-1]);
}
}
}
int main()
{
cin>>n>>m;
for(int i=1;i<=n;i++)
cin>>dp[i][0];
init();
for(int i=1;i<=n-m+1;i++)
{
int k=log2(m);
cout<<min(dp[i][k],dp[i+m-(1<<k)][k])<<endl;
}
//cout << "Hello world!" << endl;
return 0;
}
#include <iostream>
#include<cmath>
#include<cstring>
#include<cstdio>
using namespace std;
const int maxn=1e6+10;
int n,m;
int dp[maxn][25];
inline int read()
{
int 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-48;ch=getchar();}
return x*f;
}
void init()
{
int i,j;
for(j=1;j<=21;j++)
for(i=1;i+(1<<j)-1<=n;i++)
dp[i][j]=max(dp[i][j-1],dp[i+(1<<(j-1))][j-1]);
}
int main()
{
n=read();
m=read();
for(int i=1;i<=n;i++)
dp[i][0]=read();
init();
while(m--)
{
int x,y;
//cin>>x>>y;
x=read();y=read();
int len=y-x+1;
int k=log2(len);
int ans=max(dp[x][k],dp[x+len-(1<<k)][k]);
printf("%d\n",ans);
}
//cout << "Hello world!" << endl;
return 0;
}
#include <iostream>
#include<cmath>
#include<cstring>
#include<cstdio>
using namespace std;
const int maxn=1e6+10;
int n,m;
int dp[maxn][25],fm[maxn][25];
//快读
inline int read()
{
int 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-48;ch=getchar();}
return x*f;
}
//初始化
void init()
{
int i,j;
for(j=1;j<=21;j++)
for(i=1;i+(1<<j)-1<=n;i++)
{
dp[i][j]=max(dp[i][j-1],dp[i+(1<<(j-1))][j-1]);
fm[i][j]=min(fm[i][j-1],fm[i+(1<<(j-1))][j-1]);
}
}
int main()
{
n=read();
m=read();
for(int i=1;i<=n;i++)
{
dp[i][0]=read();
fm[i][0]=dp[i][0];
}
init();
while(m--)
{
int x,y;
//cin>>x>>y;
x=read();y=read();
int len=y-x+1;
int k=log2(len);
int ans=max(dp[x][k],dp[x+len-(1<<k)][k])-min(fm[x][k],fm[x+len-(1<<k)][k]);
printf("%d\n",ans);
}
//cout << "Hello world!" << endl;
return 0;
}