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\begin {array}{c} \mathfrak {One Problem Is Difficult} \\\\ \mathfrak {Because You Don't Know} \\\\ \mathfrak {Why It Is Diffucult} \end {array}

P3567 [POI2014]KUR-Couriers

P3567 [POI2014]KUR-Couriers

建立 $n$ 个值域线段树,表示 $1\ldots n$ 的出现的值域,由于每次只增加了一个数,为了节省空间就可以可持久化。查询时,$r$ 和 $l - 1$ 两棵线段树同时递归,做差几位中间的数。如果一个数出现次数大于一半,那么它所在的每一个区间都满足。

查看代码
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#include <cstdio>
using namespace std;
const int N = 5e5 + 10, M = 1e7 + 10;
struct Node
{
int l, r, s;
} tr[M];
int idx;
int n, m, w[N], rt[N];
void pushup(int x)
{
tr[x].s = tr[tr[x].l].s + tr[tr[x].r].s;
}
void modify(int &x, int l, int r, int t)
{
tr[++idx] = tr[x];
x = idx;
if (l == r)
{
tr[x].s++;
return;
}
int mid = l + r >> 1;
if (t <= mid)
modify(tr[x].l, l, mid, t);
else
modify(tr[x].r, mid + 1, r, t);
pushup(x);
}
int query(int x1, int x2, int l, int r, int k)
{
if (l == r)
return l;
int mid = l + r >> 1;
if (tr[tr[x2].l].s - tr[tr[x1].l].s > k)
return query(tr[x1].l, tr[x2].l, l, mid, k);
else if (tr[tr[x2].r].s - tr[tr[x1].r].s > k)
return query(tr[x1].r, tr[x2].r, mid + 1, r, k);
return 0;
}
int main()
{
scanf("%d%d", &n, &m);
for (int i = 1; i <= n; i++)
scanf("%d", &w[i]);
for (int i = 1; i <= n; i++)
{
rt[i] = rt[i - 1];
modify(rt[i], 1, n, w[i]);
}
for (int l, r; m; m--)
{
scanf("%d%d", &l, &r);
printf("%d\n", query(rt[l - 1], rt[r], 1, n, r - l + 1 >> 1));
}
return 0;
}