线段树二分及区间mex
作者：互联网
仓鼠的鸡蛋
思路：
把线段树权值全部置为m，维护区间最大值，每次优先找满足条件的左边的区间
const int N = 500010;
int cnt[N];
int a[N];
int n, m, k;
int tr[N << 2];
int M = 1;
void pushup(int p) {
tr[p] = max(tr[p << 1] , tr[p << 1 | 1]);
}
void build(int n) {
for (M = 1; M <= n + 5; M <<= 1);
for (int i = 1; i <= n; i++) tr[i + M] = m;
for (int i = M; i; i--) {
pushup(i);
}
}
void modify(int p, int v) {
int pos=p;
p = p + M;
int t = tr[p];
t += v;
if (++cnt[pos] == k) t = 0;
t = max(t, 0LL);
tr[p] = t;
for (p >>= 1; p; p >>= 1) pushup(p);
}
int query(int l, int r) {
int ans = -1;
for (l += M - 1, r += M + 1; l ^ r ^ 1; l >>= 1, r >>= 1) {
if (~l & 1) ans = max(ans, tr[l ^ 1]);
if (r & 1) ans = max(ans, tr[r ^ 1]);
}
return ans;
}
int find(int x, int l, int r) {
if (l == r) {
modify(l, -x);
return l;
}
int mid = l + r >> 1;
if (query(l, mid) >= x) return find(x, l, mid);
else if (query(mid + 1, r) >= x) return find(x, mid + 1, r);
return -1;
}
void solve(int Case) {
scanf("%lld%lld%lld", &n, &m, &k);
for (int i = 1; i <= n; i++) cnt[i] = 0;
for (int i = 1; i <= n; i++) scanf("%lld", &a[i]);
build(n);
for (int i = 1; i <= n; i++) {
printf("%lld\n", find(a[i], 1, n));
}
}
区间mex离线做法
思路：
只有0~n会对mex产生影响，使用线段树维护每个ai对应的最新的下标，区间维护整个区间的最小下标，
按照左端点排序，把小于r的点都放进去，然后对l进行二分，如果出现的最新的下标小于l，那么这个数字一定是mex的带选项，
ps:线段树初始化每个点为-1
代码：
const int N = 200010;
int a[N];
struct tree {
int l, r;
int mpos;
};
struct Segment_Tree {
tree tr[N << 2];
int pos[N];
#define ls(x) x<<1
#define rs(x) x<<1|1
void pushup(tree &p, tree &l, tree &r) {
p.mpos = min(l.mpos, r.mpos);
}
void pushup(int p) {
pushup(tr[p], tr[ls(p)], tr[rs(p)]);
}
void build(int p, int l, int r) {
if (l == r) {
tr[p] = {l, r, -1};
pos[l] = p;
}
else {
tr[p] = {l, r};
int mid = l + r >> 1;
build(ls(p), l, mid);
build(rs(p), mid + 1, r);
pushup(p);
}
}
void modify1(int p, int x, int y) {
p = pos[x];
tr[p] = {x, x, y};
for (; p >>= 1;) pushup(p);
}
tree query(int p, int l, int r) {
if (tr[p].l >= l and tr[p].r <= r) return tr[p];
int mid = tr[p].l + tr[p].r >> 1;
if (r <= mid) return query(ls(p), l, r);
else if (l > mid) return query(rs(p), l, r);
else {
tree ret;
auto left = query(ls(p), l, r);
auto right = query(rs(p), l, r);
pushup(ret, left, right);
return ret;
}
}
int find(int p, int l, int r, int x) {
if (l == r) {
return l;
}
int mid = l + r >> 1;
if (query(p, l, mid).mpos < x) {
return find(p << 1, l, mid, x);
} else return find(p << 1 | 1, mid + 1, r, x);
}
};
Segment_Tree ST;
struct T {
int l, r, mex, ans, id;
bool operator<(const T &t) const {
return r < t.r;
}
} Q[N];
int c[N];
#define lowbit(x) x&(-x)
void add(int x, int y) {
for (; x < N; x += lowbit(x)) c[x] += y;
}
int ask(int x) {
int res = 0;
for (; x; x -= lowbit(x)) res += c[x];
return res;
}
void solve(int Case) {
int n = read(), m = read();
for (int i = 1; i <= n; i++) {
a[i] = read();
}
for (int i = 1; i <= m; i++) {
auto&[l, r, _, __, id] = Q[i];
l = read(), r = read();
id = i;
}
ST.build(1, 0, n);
sort(Q + 1, Q + 1 + m);
int p = 1;
for (int i = 1; i <= m; i++) {
auto&[l, r, mex, __, id] = Q[i];
//cout << l << ' ' << r << nline;
for (int i = p; i <= r; i++) {
if (a[i] <= n) {
ST.modify1(1, a[i], i);
}
}
p = r + 1;
mex = ST.find(1, 0, n, l);
}
sort(Q + 1, Q + 1 + m, [](T a, T b) {return a.id < b.id;});
for (int i = 1; i <= m; i++) {
printf("%lld\n", Q[i].mex);
}
}
little w and Discretization
思路：
离散化之后答案一定是这段大于区间的mex(从1开始算)的数的个数，求mex直接套上题，区间改为1~n+1，求大于区间内某个数的个数，同样是离线，使用线段树或树状数组
代码：
const int N = 300010;
int a[N];
struct tree {
int l, r;
int mpos;
};
struct Segment_Tree {
tree tr[N << 2];
int pos[N];
#define ls(x) x<<1
#define rs(x) x<<1|1
void pushup(tree &p, tree &l, tree &r) {
p.mpos = min(l.mpos, r.mpos);
}
void pushup(int p) {
pushup(tr[p], tr[ls(p)], tr[rs(p)]);
}
void build(int p, int l, int r) {
if (l == r) {
tr[p] = {l, r, -1};
pos[l] = p;
}
else {
tr[p] = {l, r};
int mid = l + r >> 1;
build(ls(p), l, mid);
build(rs(p), mid + 1, r);
pushup(p);
}
}
void modify1(int p, int x, int y) {
p = pos[x];
tr[p] = {x, x, y};
for (; p >>= 1;) pushup(p);
}
tree query(int p, int l, int r) {
if (tr[p].l >= l and tr[p].r <= r) return tr[p];
int mid = tr[p].l + tr[p].r >> 1;
if (r <= mid) return query(ls(p), l, r);
else if (l > mid) return query(rs(p), l, r);
else {
tree ret;
auto left = query(ls(p), l, r);
auto right = query(rs(p), l, r);
pushup(ret, left, right);
return ret;
}
}
int find(int p, int l, int r, int x) {
if (l == r) {
return l;
}
int mid = l + r >> 1;
if (tr[p << 1].mpos < x) {
return find(p << 1, l, mid, x);
} else return find(p << 1 | 1, mid + 1, r, x);
}
};
Segment_Tree ST;
struct T {
int l, r, mex, ans, id;
bool operator<(const T &t) const {
return r < t.r;
}
} Q[N];
int c[N];
#define lowbit(x) x&(-x)
void add(int x, int y) {
for (; x < N; x += lowbit(x)) c[x] += y;
}
int ask(int x) {
int res = 0;
for (; x; x -= lowbit(x)) res += c[x];
return res;
}
void solve(int Case) {
int n = read();
for (int i = 1; i <= n; i++) {
a[i] = read();
}
int m = read();
for (int i = 1; i <= m; i++) {
auto&[l, r, _, __, id] = Q[i];
l = read(), r = read();
id = i;
}
ST.build(1, 1, n );
sort(Q + 1, Q + 1 + m);
int p = 1;
for (int i = 1; i <= m; i++) {
auto&[l, r, mex, __, id] = Q[i];
for (int i = p; i <= r; i++) {
if (a[i] <= n) {
ST.modify1(1, a[i], i);
}
}
p = r + 1;
mex = ST.find(1, 1, n, l);
}
sort(Q + 1, Q + 1 + m, [](T a, T b) {return a.id < b.id;});
using PII = pair<int, int>;
vector<PII> v(n + 1);
for (int i = 1; i <= n; i++) {
auto &[x, y] = v[i];
x = a[i], y = i;
}
sort(v.begin() + 1, v.end(), greater<PII>());
sort(Q + 1, Q + 1 + m, [](T a, T b) {return a.mex > b.mex;});
p = 1;
for (int i = 1; i <= m; i++) {
auto &[l, r, mex, ans, id] = Q[i];
while (p <= n and v[p].first > mex) {
add(v[p].second, 1);
p++;
}
ans = (ask(r) - ask(l - 1));
}
sort(Q + 1, Q + 1 + m, [](T a, T b) {return a.id < b.id;});
for (int i = 1; i <= m; i++) {
printf("%lld\n", Q[i].ans);
}
}
标签：二分,return,int,线段,tr,mid,query,mex 来源： https://www.cnblogs.com/koto-k/p/16204263.html