【UOJ #246】【UER #7】套路

http://uoj.ac/contest/35/problem/246

神奇!我这辈子是想不出这样的算法了。

对区间长度分类讨论:题解很好的~

我已经弱到爆了,看完题解后还想了一晚上。

题解中“利用\(r_y\)进行计算更新答案”的具体方法是记录以当前点为右端点,任意两个数的差值的最小值大于等于j的区间的左端点,记为\(pos_j\)。

就这个问题我想了一晚上啊TWT,我不滚粗谁滚粗QAQ

#include<cmath>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N = 200003;
int in() {
int k = 0; char c = getchar();
for (; c < '0' || c > '9'; c = getchar());
for (; c >= '0' && c <= '9'; c = getchar())
k = k * 10 + c - 48;
return k;
} int S, n, m, k, a[N], ans = 0, f[N], up, last[N], pos[N]; void solve_1() {
for (int i = 1; i < n; ++i) {
f[i] = abs(a[i + 1] - a[i]);
if (k <= 2) ans = max(ans, f[i]);
} for (int p = 3; p <= S; ++p)
for (int i = 1; i + p - 1 <= n; ++i) {
f[i] = min(abs(a[i + p - 1] - a[i]), min(f[i], f[i + 1]));
if (k <= p) ans = max(ans, f[i] * (p - 1));
}
} void solve_2() {
int lo, bi;
for (int i = 1; i <= n; ++i) {
pos[0] = max(pos[0], last[a[i]]);
for (int j = 1; j <= up; ++j) {
pos[j] = max(pos[j], pos[j - 1]);
lo = a[i] - j;
bi = a[i] + j;
if (lo >= 1) pos[j] = max(pos[j], last[lo]);
if (bi <= m) pos[j] = max(pos[j], last[bi]);
if (i - pos[j - 1] >= k)
ans = max(ans, j * (i - (pos[j - 1] + 1)));
}
last[a[i]] = i;
}
} int main() {
n = in(); m = in(); k = in();
for (int i = 1; i <= n; ++i)
a[i] = in();
S = ceil(sqrt(n)); solve_1(); up = m / S;
solve_2(); printf("%d\n", ans);
return 0;
}
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