【BZOJ3339】Rmq Problem【离线】【线段树】【mex】

奇怪的线段树姿势get。

看了hzwer的题解，发现这个奇怪的姿势。

权值也是标记...

/* Footprints In The Blood Soaked Snow */
#include <cstdio>
#include <algorithm>

using namespace std;

const int maxn = 200005, maxm = maxn, inf = 0x3f3f3f3f;

int n, m, num[maxn], next[maxn], last[maxn], sg[maxn], ans[maxm], tr[maxn << 2];
bool vis[maxn];

struct _query {
	int id, l, r;

	bool operator < (const _query &a) const {
		return l < a.l;
	}
} q[maxm];

inline int iread() {
	int f = 1, x = 0; char ch = getchar();
	for(; ch < '0' || ch > '9'; ch = getchar()) f = ch == '-' ? -1 : 1;
	for(; ch >= '0' && ch <= '9'; ch = getchar()) x = x * 10 + ch - '0';
	return f * x;
}

inline void build(int p, int l, int r) {
	tr[p] = inf;
	if(l == r) {
		tr[p] = sg[l];
		return;
	}
	int mid = l + r >> 1;
	build(p << 1, l, mid); build(p << 1 | 1, mid + 1, r);
}

inline void pushdown(int p) {
	if(tr[p] != inf) {
		tr[p << 1] = min(tr[p], tr[p << 1]);
		tr[p << 1 | 1] = min(tr[p], tr[p << 1 | 1]);
		tr[p] = inf;
	}
}

inline void change(int p, int l, int r, int x, int y, int c) {
	if(x <= l && r <= y) {
		tr[p] = min(tr[p], c);
		return;
	}
	int mid = l + r >> 1;
	pushdown(p);
	if(x <= mid) change(p << 1, l, mid, x, y, c);
	if(y > mid) change(p << 1 | 1, mid + 1, r, x, y, c);
}

inline int query(int p, int l, int r, int x) {
	if(l == r && r == x) return tr[p];
	int mid = l + r >> 1;
	pushdown(p);
	if(x <= mid) return query(p << 1, l, mid, x);
	else return query(p << 1 | 1, mid + 1, r, x);
}

int main() {
	n = iread(); m = iread();

	for(int i = 1, k = 0; i <= n; i++) {
		vis[num[i] = iread()] = 1;
		for(; vis[k]; k++);
		sg[i] = k;
	}
	for(int i = 1; i <= m; i++) {
		int l = iread(), r = iread();
		q[i] = (_query){i, l, r};
	}
	sort(q + 1, q + 1 + m);

	build(1, 1, n);
	for(int i = n; i >= 1; i--) next[i] = last[num[i]], last[num[i]] = i;
	for(int i = 1; i <= n; i++) if(!next[i]) next[i] = n + 1;

	for(int i = 1, pos = 1; i <= m; i++) {
		for(; pos < q[i].l; pos++)
			if(pos + 1 <= next[pos] - 1)
				change(1, 1, n, pos + 1, next[pos] - 1, num[pos]);
		ans[q[i].id] = query(1, 1, n, q[i].r);
	}

	for(int i = 1; i <= m; i++) printf("%d\n", ans[i]);
	return 0;
}