HDOJ 3308 LCIS(线段树区间合并与查询)
题意:
给定n个数,可以对某一个数进行更新操作,以及对于某一个区间进行“最长连续上升子序列”查询。
思路:
和上一题hotel类似,只不过减少了一个PushDown的操作。多了对于数组中元素的判断,要细心体会。
#include <cstdio>
#include <algorithm>
using namespace std;
#define lhs l, m, rt << 1
#define rhs m + 1, r, rt << 1 | 1
const int maxn = 100010;
int sum[maxn << 2], lsum[maxn << 2], rsum[maxn << 2];
int arr[maxn];
void PushUp(int l, int r, int rt)
{
lsum[rt] = lsum[rt << 1];
rsum[rt] = rsum[rt << 1 | 1];
int m = (l + r) >> 1;
sum[rt] = max(sum[rt << 1], sum[rt << 1 | 1]);
if (arr[m] < arr[m + 1])
{
if (lsum[rt] == m - l + 1)
lsum[rt] += lsum[rt << 1 | 1];
if (rsum[rt] == r - m)
rsum[rt] += rsum[rt << 1];
sum[rt] = max(sum[rt], rsum[rt << 1] + lsum[rt << 1 | 1]);
}
}
void Build(int l, int r, int rt)
{
if (l == r)
{
sum[rt] = 1;
lsum[rt] = rsum[rt] = 1;
return ;
}
int m = (l + r) >> 1;
Build(lhs); Build(rhs);
PushUp(l, r, rt);
}
void Update(int p, int value, int l, int r, int rt)
{
if (l == r)
{
arr[p] = value;
return ;
}
int m = (l + r) >> 1;
if (p <= m)
Update(p, value, lhs);
else
Update(p, value, rhs);
PushUp(l, r, rt);
}
int Query(int beg, int end, int l, int r, int rt)
{
if (beg <= l && r <= end)
return sum[rt];
int m = (l + r) >> 1;
int ret = 0;
if (beg <= m)
ret = max(ret, Query(beg, end, lhs));
if (end > m)
ret = max(ret, Query(beg, end, rhs));
if (arr[m] < arr[m + 1])
if (beg <= m && m < end)
ret = max(ret, min(m - beg + 1, rsum[rt << 1]) + min(end - m, lsum[rt << 1 | 1]));
return ret;
}
int main()
{
int cases;
scanf("%d", &cases);
while (cases--)
{
int n, m;
scanf("%d %d", &n, &m);
for (int i = 0; i < n; ++i)
scanf("%d", &arr[i]);
Build(0, n - 1, 1);
for (int i = 0; i < m; ++i)
{
char op[4];
int a, b;
scanf("%s %d %d", op, &a, &b);
if (op[0] == 'U')
Update(a, b, 0, n - 1, 1);
else if (op[0] == 'Q')
printf("%d\n", Query(a, b, 0, n - 1, 1));
}
}
return 0;
}
-------------------------------------------------------
kedebug
Department of Computer Science and Engineering,
Shanghai Jiao Tong University
E-mail: kedebug0@gmail.com
GitHub: http://github.com/kedebug
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