HDOJ-1806 ( Frequent values ) 线段树区间合并

http://acm.hdu.edu.cn/showproblem.php?pid=1806

线段树维护区间出现频率最高的出现次数。为了维护上者，需要维护线段前后缀的出现次数，当和其他线段在端点处的字符相等时，要做合并处理。

#include <iostream>
#include <cstdlib>
#include <vector>
#include <string>
#include <cmath>
#include <algorithm>
#include <map>
using namespace std;
int n, m;
const int len = 100005;
int num[len];
struct node
{
    int l, r;
    int ll, rr;
    int v;
};
node tr[len << 2];
void pushUp(int rt)
{
    tr[rt].v = max(tr[rt << 1].v, tr[rt << 1 | 1].v);
    if (num[tr[rt << 1].r] == num[tr[rt << 1 | 1].l])
        tr[rt].v = max(tr[rt].v, tr[rt << 1].rr + tr[rt << 1 | 1].ll);
    tr[rt].ll = tr[rt << 1].ll;
    tr[rt].rr = tr[rt << 1 | 1].rr;
    if (num[tr[rt].l] == num[tr[rt << 1 | 1].l])
        tr[rt].ll += tr[rt << 1 | 1].ll;
    if (num[tr[rt].r] == num[tr[rt << 1].r])
        tr[rt].rr += tr[rt << 1].rr;
}
void build(int l, int r, int rt)
{
    tr[rt].l = l;
    tr[rt].r = r;
    if (l == r)
    {
        tr[rt].v = 1;
        tr[rt].ll = tr[rt].rr = 1;
        return;
    }
    int mid = (l + r) >> 1;
    build(l, mid, rt << 1);
    build(mid + 1, r, rt << 1|1);
    pushUp(rt);
}
int query(int l, int r, int rt)
{
    if (l <= tr[rt].l&&tr[rt].r <= r)
        return tr[rt].v;
    int mid = (tr[rt].l + tr[rt].r) >> 1;
    int lr=-1, rr=-1;
    if (l <= mid)
        lr = query(l, r, rt << 1);
    if (r > mid)
        rr = query(l, r, rt << 1 | 1);
    int ans = max(lr, rr);
    if (lr != -1 && rr != -1&&num[mid]==num[mid+1])
    {
        ans = max(ans, min(tr[rt << 1].rr,mid-l+1) + min(r-mid,tr[rt << 1 | 1].ll));
    }
    //if (ans == 4) cout << tr[rt].l << ' ' << tr[rt].r<<' '<<lr<<' '<<rr<< endl;
    return ans;
}
int main()
{
    while (scanf("%d", &n) != EOF)
    {
        if (n == 0) break;
        scanf("%d", &m);
        for (int i = 1; i <= n; i++)
            scanf("%d", &num[i]);
        build(1, n, 1);
        while (m--)
        {
            int l, r;
            scanf("%d%d", &l , &r);
            printf("%d\n", query(l, r, 1));
        }
    }
    return 0;
}