思路
- 询问的是区间[L, R],完美序列的开始下标 >= L
- 二分查找序列中满足开始下标大于L的第一个位置x
- ST算法求解[x, R]的最大长度 y
- x - L 和 y 的最大值为结果
代码
#include <cstdio>
#define max(a,b) ((a) > (b) ? (a) : (b))
const int MAX = 1000000;
int arr[200005], last[2000006], st[200005];
int f[200005][21], log[200005], rec[200005];
int res;
inline int read() {
int s = 0, f = 1;
char ch = getchar();
while (ch < '0' || ch > '9') {
if (ch == '-') f = -1;
ch = getchar();
}
while (ch >= '0' && ch <= '9') s = s * 10 + ch - '0', ch = getchar();
return s * f;
}
int n, m;
// 找出可以应用st算法的第一个值
int find(int l, int r) {
if (st[l] >= l) return l;
if (st[r] < l) return r + 1;
int ans = 0, L = l, R = r, mid;
while (l <= r) {
mid = (l + r) >> 1;
if (st[mid] >= L) ans = mid, r = mid - 1;
else l = mid + 1;
}
return ans;
}
int query(int l, int r) {
int x = log[r - l + 1];
return max(f[l][x], f[r - (1 << x) + 1][x]);
}
int main() {
n = read(), m = read();
log[0] = -1;
for (int i = 1; i <= n; ++i) {
arr[i] = read();
log[i] = log[i >> 1] + 1;
st[i] = max(st[i - 1], last[arr[i] + MAX] + 1);
rec[i] = i - st[i] + 1;
f[i][0] = rec[i];
last[arr[i] + MAX] = i;
}
for (int i = 1; i <= 20; ++i) {
for (int j = 1; j + (1 << i) - 1 <= n; ++j) {
f[j][i] = max(f[j][i - 1], f[j + (1 << (i - 1))][i - 1]);
}
}
for (int i = 1, a, b, c, d, e; i <= m; ++i) {
a = read(), b = read();
res = 0, a++, b++;
c = find(a, b);
if (c > a) res = c - a;
if (c <= b) res = max(res, query(c, b));
printf("%d\n", res);
}
return 0;
}