区间专题

一. 区间最大最小值问题

　　1. RMQ

int mx[N][20];                      //最多能保存524288的长度
int RMQ(int l, int r)   {
    int k = 0;  while (1<<(k+1) <= r - l + 1)   k++;        //令k为满足1<<k <= r-l+1的最大整数
    return max (mx[l][k], mx[r-(1<<k)+1][k]);               //区间最左边1<<k长度的最大值和最右边1<<k长度的最大值,可能有重叠
}

int main(void)    {
        for (int i=1; i<=n; ++i)    {
            scanf ("%d", &mx[i][0]);
        }
        for (int j=1; (1<<j)<=n; ++j)    {
            for (int i=1; i+(1<<j)-1<=n; ++i)   {
                mx[i][j] = max (mx[i][j-1], mx[i+(1<<(j-1))][j-1]);     //从i开始，长度1<<j的区间的最大值
            }
        }
        printf ("%d
", RMQ (ql, qr));
｝

　　2. Segment_Tree

struct ST    {
    int mx[N<<2];
    void push_up(int rt)    {
        mx[rt] = max (mx[rt<<1], mx[rt<<1|1]);
    }
    void build(int l, int r, int rt)    {
        if (l == r)    {
            scanf ("%d", &mx[rt]);    return ;
        }
        int mid = (l + r) >> 1;
        build (lson);　　build (rson);
        push_up (rt);
    }
    int query(int ql, int qr, int l, int r, int rt)    {
        if (ql <= l && r <= qr)    {
            return mx[rt];
        }
        int mid = (l + r) >> 1, ret = 0;
        if (ql <= mid)    ret = max (ret, query (ql, qr, lson));
        if (qr > mid)    ret = max (ret, query (ql, qr, rson));
        return ret;
    }
}st;