POJ 2823

很慢的线段树，不过可以作为最简单的模板。代码风格学的Not Only Success。C++过,G++不过。

#include <cstdio>
#include <climits>

#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1

const int MAX = 1000005;
int minarray[MAX<<2];
int maxarray[MAX<<2];
int n,k;

int min(int a,int b)
{
    return a>b?b:a;
}

int max(int a,int b)
{
    return a>b?a:b;
}

void build(int l,int r,int rt)
{
    if(l == r)
    {
        int temp;
        scanf("%d",&temp);
        minarray[rt] = temp;
        maxarray[rt] = temp;
        return;
    }

    int m = (l+r)>>1;
    build(lson);
    build(rson);

    minarray[rt] = min(minarray[rt<<1], minarray[rt<<1|1]);
    maxarray[rt] = max(maxarray[rt<<1], maxarray[rt<<1|1]);
}

int querymin(int L,int R, int l,int r,int rt)
{
    if(L <= l && r <= R)
    {
        return minarray[rt];
    }

    int m = (l+r)>>1;
    int ans = INT_MAX;
    if(L <= m) ans = min(ans, querymin(L, R, lson));
    if(m < R) ans = min(ans, querymin(L, R, rson));

    return ans;
}

int querymax(int L,int R, int l,int r,int rt)
{
    if(L <= l && r <= R)
    {
        return maxarray[rt];
    }

    int m = (l+r)>>1;
    int ans = INT_MIN;
    if(L <= m) ans = max(ans, querymax(L, R, lson));
    if(m < R) ans = max(ans, querymax(L, R, rson));

    return ans;
}


int main()
{
    scanf("%d %d", &n, &k);
    build(1,n,1);

    for(int i = 1; i <= n-k; i++)
    {
        printf("%d ", querymin(i, i+k-1, 1, n, 1));
    }
    printf("%d\n", querymin(n-k+1, n, 1, n, 1));


    for(int i = 1; i <= n-k; i++)
    {
        printf("%d ", querymax(i, i+k-1, 1, n, 1));
    }
    printf("%d\n", querymax(n-k+1, n, 1, n, 1));
}