[ACM] RMQ板子题的多种解法（线段树，ST表，数列分块）

P2251 质量检测

线段树

#include <bits/stdc++.h>
#define ls ( rt<<1 )
#define rs ( rt<<1|1 )
#define Mid ( T[rt].l + T[rt].r >> 1 )
#define L ( T[rt].l )
#define R ( T[rt].r )
#define inf 0x3f3f3f3f
using namespace std;

int n,m;
const int N = 1e6+10;
struct Node
{
    int l,r;
    int mi;
}T[N];

int a[N];

void push_up(int rt)
{
    T[rt].mi = min( T[ls].mi , T[rs].mi );
}

void build(int rt,int l,int r)
{
    T[rt] = {l,r,0};
    if(l == r){
        T[rt].mi = a[l];
        return ;
    }
    build( ls,l,Mid),build(rs,Mid+1,r);
    push_up(rt);
}

int query(int rt,int l,int r)
{
    if( l <= L && r >= R ){
        return T[rt].mi;
    }
    int al = inf,ar = inf;
    if( l <= Mid ) al = query(ls,l,r);
    if( r > Mid ) ar = query(rs,l,r);
    return min(al,ar);
}


int main()
{
    scanf("%d%d",&n,&m);
    for(int i = 1 ; i <= n ; i++)   scanf("%d",&a[i]);
    build(1,1,n);
    for(int i = 1 ; i + m - 1<= n ; i++){
        printf("%d
",query(1,i,i+m-1) );
    }

}

ST表

#include <bits/stdc++.h>
using namespace std;

int n,m;
const int N = 1e6+10;
int a[N][21];

inline void pre()
{
    for(int j = 1; j <= 21; j ++)
        for(int i = 1; i + (1<<j) - 1 <= n ; i++)
            a[i][j] = min( a[i][j-1],a[ i + (1 << (j-1)) ][j-1] );
}

inline int query(int l,int r)
{
    int k = log2(r-l+1);
    return min( a[l][k], a[ r-(1<<k) + 1 ][k] );
}

int main()
{
    scanf("%d%d",&n,&m);
    for(int i = 1; i <= n ; i++) scanf("%d",&a[i]);

    pre();

    for(int i = 1 ; i + m - 1<= n ; i++){
         printf("%d
",query(i,i+m-1) );
    }

    return 0;
}

分块

#include <bits/stdc++.h>
using namespace std;

const int N = 1e6+10;
int mi[N];
int a[N];
int id[N];
int n,m,block;

int query(int l,int r)
{
    int sid = id[l],eid = id[r];
    int ans = 0x3f3f3f3f;
    if( sid == eid ){
        for(int i = l; i <= r; i++) ans = min(ans, a[i] );
        return ans;
    }

    for(int i = l ; id[i] == sid ; i++) ans = min(ans,a[i]);
    for(int i = sid + 1; i < eid ; i++) ans = min(ans,mi[i]);
    for(int i = (eid-1)*block + 1 ; i <=r ; i++ ) ans = min(ans,a[i]);
    return ans ;
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);

    cin >> n >> m ;
    block = sqrt(n);
    int len = (n-1)/block +1 ;
    for(int i = 1 ; i <= len; i ++){
        mi[i] = 0x3f3f3f3f;
    }

    for(int i = 1; i <= n ; i++){
        cin >> a[i];
        id[i] = ( i - 1 ) / block + 1;
        mi[ id[i] ] = min( mi[id[i]] , a[i] );
    }

    for(int i = 1; i + m - 1 <=  n; i ++){
        cout <<query(i,i+m-1) << endl;
    }

    return 0;
}