poj-2823 -- Sliding Window 区间最大值最小值 RMQ 与线段树

题意：一个有n个元素的数组num[]，求所有num[i]~num[i+k]的最小值和最大值。

RMQ 算法小心空间，很容易被卡到

可以用线段树做,每个结点保存该线段的最大/小值...然后简单查询最大最小值即可

RMQ：

//Accepted    8008K    6110MS    C++    1577B 超卡空间
#include <cstdio>
#include <string>
#include <cmath>
#include <algorithm>
using namespace std;
const int MAX = 1000001;
int n,q;
int st[MAX];
int val[MAX];
inline int minn(int a,int b){ return a>b?b:a; }
inline int maxx(int a,int b){ return a>b?a:b; }
void make_st_max()
{
    int i,j,k;
    for( i=1; i<=n; i++)
          st[i] = val[i];    //实质是: dp[i][0]
    k = log(double(q)) / log(2.0);
    for( j=1; j<=k; j++)
        for( i=1; i+(1<<j)-1 <= n; i++)
                            st[i] = maxx( st[i] , st[i+(1<<(j-1))]);  
//实质是: max( dp[i][j-1] , dp[i+(1<<(j-1))][j-1]); 这里,因为有用到dp[i][log(double(K)) / log(2.0)]的值,所以用滚动数组就可以了
}
void make_st_min()
{
    int i,j,k;
    for( i=1; i<=n; i++)
          st[i] = val[i];    //实质是: dp[i][0]
    k = log(double(q)) / log(2.0);
    for( j=1; j<=k; j++)
        for( i=1; i+(1<<j)-1 <= n; i++)
                            st[i] = minn( st[i] , st[i+(1<<(j-1))]);  
        //实质是: max( dp[i][j-1] , dp[i+(1<<(j-1))][j-1]); 这里,因为有用到dp[i][log(double(K)) / log(2.0)]的值,所以用滚动数组就可以了
}

int rmq(int a,int b,int flag)
{    
    int k;
    k=floor(log((double)(b-a+1))/log(2.0));
    if(flag > 0)
    {
        return  maxx(st[a], st[b-(1<<k)+1]);
    }
    else
    {
        return minn(st[a], st[b-(1<<k)+1]);
    }
}

int main(void)
{
    int i;
    while(scanf("%d%d",&n,&q)!=EOF)
    {
        for(i=1;i<=n;i++)
        {
            scanf("%d",&val[i]);
        }
        make_st_min();
        for(i=1;i<=n-q+1;i++)
        {
            printf("%d ",rmq(i,i+q-1,-1));
        }
        printf("\n");
        make_st_max();
        for(i=1;i<=n-q+1;i++)
        {
            printf("%d ",rmq(i,i+q-1,1));
        }
        printf("\n");
    }
    return 0;
}

线段树：

//Accepted    16596K    9204MS    C++    1487B
#include <stdio.h>
#include <string.h>
#define  inf 0x7fffffff
#define lson l , m , rt << 1
#define rson m + 1 , r , rt << 1 | 1
const int maxn = 1000001;

int st_min[maxn<<2],st_max[maxn<<2];

inline int minn(int a,int b) { return a>b?b:a; }
inline int maxx(int a,int b) { return a>b?a:b; }
void PushUP(int rt)
{
    st_min[rt] = minn(st_min[rt<<1],st_min[rt<<1|1]);
    st_max[rt] = maxx(st_max[rt<<1],st_max[rt<<1|1]);
}
void build(int l,int r,int rt) {
    if (l == r)
    {
        scanf("%d",&st_min[rt]);
        st_max[rt] = st_min[rt];
        return ;
    }
    int m = (l + r) >> 1;
    build(lson);
    build(rson);
    PushUP(rt);
}

int query_min(int L,int R,int l,int r,int rt) 
{
    if (L <= l && r <= R) {
        return st_min[rt];
    }
    int m = (l + r) >> 1;
    int ret1 = inf,ret2 = inf;
    if (L <= m) ret1 = query_min(L , R , lson);
    if (R > m) ret2 = query_min(L , R , rson);
    return minn(ret1,ret2);
}
int query_max(int L,int R,int l,int r,int rt) 
{
    if (L <= l && r <= R) {
        return st_max[rt];
    }
    int m = (l + r) >> 1;
    int ret1 = -inf,ret2 = -inf;
    if (L <= m) ret1 = query_max(L , R , lson);
    if (R > m) ret2 = query_max(L , R , rson);
    return maxx(ret1,ret2);
}

int main(void)
{
    int n,m,i;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        build(1 , n , 1);
        for(i=1;i<=n-m+1;i++)
        {
            printf("%d ",query_min(i,i+m-1,1,n,1));
        }
        printf("\n");
        for(i=1;i<=n-m+1;i++)
        {
            printf("%d ",query_max(i,i+m-1,1,n,1));
        }
        printf("\n");
    }
    return 0;
}