数据结构 Week 1 --- 从线段树到主席树
作者：互联网
线段树的静态区间查询
- 区间里大于x的最左边的位置：
看左区间的最大值是否大于x，决定是否往左区间找
- 区间最大连续和：
维护区间左连续最大和，右连续最大和，最大连续和
- 求整个区间中大于k的数之和：
权值线段树，然后求区间和
- 在排好序的数组里查询区间众数的个数：
维护区间左连续相同数的个数，区间右连续相同数的个数，区间连续相同数的个数
有序区间的区间众数个数代码：
struct NODE {
int l, r, mid;
int lc, rc, mc;
}tree[MAXN*4];
void build(int pos, int l, int r) {
tree[pos].l = l; tree[pos].r = r; tree[pos].mid = l + r >> 1;
if (l == r) {
tree[pos].lc = tree[pos].rc = tree[pos].mc = 1;
return;
}
int mid = l + r >> 1;
build(pos << 1, l, mid);
build(pos << 1 | 1, mid + 1, r);
if (a[l] == a[mid + 1]) tree[pos].lc = tree[pos << 1].lc + tree[pos << 1 | 1].lc;
else tree[pos].lc = tree[pos << 1].lc;
if (a[mid] == a[r]) tree[pos].rc = tree[pos << 1 | 1].rc + tree[pos << 1].rc;
else tree[pos].rc = tree[pos << 1 | 1].rc;
tree[pos].mc = max(tree[pos << 1].mc, tree[pos << 1 | 1].mc);
if (a[mid] == a[mid + 1]) tree[pos].mc = max(tree[pos].mc, tree[pos << 1].rc + tree[pos << 1 | 1].lc);
}
int Q_l(int pos, int l, int r) {
if (tree[pos].l == l && tree[pos].r == r) return tree[pos].lc;
int mid = tree[pos].mid;
if (r <= mid) return Q_l(pos << 1, l, r);
else if (l > mid) return Q_l(pos << 1 | 1, l, r);
else {
if (a[l] == a[mid + 1]) return Q_l(pos << 1, l, mid) + Q_l(pos << 1 | 1, mid + 1, r);
else return Q_l(pos << 1, l, mid);
}
}
int Q_r(int pos, int l, int r) {
if (tree[pos].l == l && tree[pos].r == r) return tree[pos].rc;
int mid = tree[pos].mid;
if (r <= mid) return Q_r(pos << 1, l, r);
else if (l > mid) return Q_r(pos << 1 | 1, l, r);
else {
if (a[mid] == a[r]) return Q_r(pos << 1, l, mid) + Q_r(pos << 1 | 1, mid + 1, r);
else return Q_r(pos<<1|1,mid+1,r);
}
}
int Q(int pos, int l, int r) {
if (tree[pos].l == l && tree[pos].r == r) return tree[pos].mc;
int mid = tree[pos].mid;
if (r <= mid) return Q(pos << 1, l, r);
else if (l > mid) return Q(pos << 1 | 1, l, r);
else {
int res = max(Q(pos << 1, l, mid), Q(pos << 1 | 1, mid + 1, r));
if (a[mid] == a[mid + 1]) res = max(res, Q_r(pos << 1, l, mid) + Q_l(pos << 1 | 1, mid + 1, r));
return res;
}
}
线段树的区间修改和懒标记
- 扫描线模板题中维护区间里非0数的个数：
记录区间被矩形完全覆盖的矩形个数，查询时，如果区间被某个矩形完全覆盖，非0数=区间长度，否则非0数=左区间非0数+右区间非0数
因为查询都是查询[1,n]区间的非0数，所以不需要懒标记
- 以函数映射为懒标记的线段树：
和一般懒标记一样，写码时要小心，lazy_tag别忘了处理
- 宾馆住户题，找到宾馆里最靠左的有连续x个空房的房间并入住，宾馆区间[x,y]退房：
维护区间左连续空房数，右连续空房数，最大连续空房数，还有区间修改需要的懒标记
宾馆住户题代码：
#include<iostream>
#include<algorithm>
#include<cstdio>
using namespace std;
const int MAXN = 1e5 + 7;
struct NODE {
int l, r, mid, len;
int lazy, lc, rc, mc;
}tree[MAXN * 4];
void build(int pos, int l, int r) {
tree[pos].l = l; tree[pos].r = r; tree[pos].mid = l + r >> 1; tree[pos].len = r - l + 1;
tree[pos].lazy = 2;
tree[pos].lc = tree[pos].rc = tree[pos].mc = tree[pos].len;
if (l == r) return;
int mid = l + r >> 1;
build(pos << 1, l, mid);
build(pos << 1 | 1, mid + 1, r);
}
void pd(int pos) {
int lazy = tree[pos].lazy;
tree[pos << 1].lazy = tree[pos << 1 | 1].lazy = lazy;
if (lazy == 1) {
tree[pos << 1].lc = tree[pos << 1].rc = tree[pos << 1].mc = 0;
tree[pos << 1 | 1].lc = tree[pos << 1 | 1].rc = tree[pos << 1 | 1].mc = 0;
}
else {
tree[pos << 1].lc = tree[pos << 1].rc = tree[pos << 1].mc = tree[pos << 1].len;
tree[pos << 1 | 1].lc = tree[pos << 1 | 1].rc = tree[pos << 1 | 1].mc = tree[pos << 1 | 1].len;
}
tree[pos].lazy = 2;
}
void update(int pos) {
if (tree[pos << 1].lc == tree[pos<<1].len) tree[pos].lc = tree[pos << 1].len + tree[pos << 1 | 1].lc;
else tree[pos].lc = tree[pos << 1].lc;
if (tree[pos << 1 | 1].rc == tree[pos << 1 | 1].len) tree[pos].rc = tree[pos << 1 | 1].len + tree[pos << 1].rc;
else tree[pos].rc = tree[pos << 1 | 1].rc;
tree[pos].mc = max(tree[pos << 1].mc, tree[pos << 1 | 1].mc);
tree[pos].mc = max(tree[pos].mc, tree[pos << 1].rc + tree[pos << 1 | 1].lc);
}
void CHANGE(int pos, int l, int r, int k) {
if (tree[pos].l == l && tree[pos].r == r) {
tree[pos].lazy = k;
if (k) {
tree[pos].lc = tree[pos].rc = tree[pos].mc = 0;
}
else {
tree[pos].lc = tree[pos].rc = tree[pos].mc = tree[pos].len;
}
return;
}
if (tree[pos].lazy != 2) pd(pos);
int mid = tree[pos].mid;
if (r <= mid) CHANGE(pos << 1, l, r, k);
else if (l > mid) CHANGE(pos << 1 | 1, l, r, k);
else {
CHANGE(pos << 1, l, mid, k);
CHANGE(pos << 1 | 1, mid + 1, r, k);
}
update(pos);
}
int Q(int pos, int x) {
if (tree[pos].mc < x) return 0;
if (tree[pos].lazy != 2) pd(pos);
if (tree[pos << 1].mc >= x) return Q(pos << 1, x);
if (tree[pos << 1].rc + tree[pos << 1 | 1].lc >= x) return tree[pos << 1].r - tree[pos << 1].rc + 1;
else return Q(pos << 1|1, x);
}
int main()
{
int n, m, op, x, y;
cin >> n >> m;
build(1, 1, n);
while (m--) {
scanf("%d%d", &op, &x);
if (op == 2) {
scanf("%d", &y);
CHANGE(1, x, x + y - 1, 0);
}
else {
int pp = Q(1, x);
printf("%d\n", pp);
if (pp) CHANGE(1, pp, pp + x - 1, 1);
}
}
return 0;
}
标签：Week,树到,return,int,tree,pos,mid,---,区间 来源： https://www.cnblogs.com/ruanbaitql/p/14714614.html