平衡树做题记录
平衡树做题记录
板子就不说了。
P2786 英语1（eng1）- 英语作文
红黑树 map 随便做，用一个 map 存下字符串对应的值，一个字符一个字符读入，然后判断， 如果不是数字并且不是字母，说明空格或者符号，处理答案。
/**
* author: TLE_Automation
* creater: 2022.9.7
**/
#include<cmath>
#include<queue>
#include<cstdio>
#include<bitset>
#include<cstring>
#include<iostream>
#include<algorithm>
#define gc getchar
using namespace std;
typedef long long ll;
const int N = 1e6 + 10;
const int mod = 998244353;
const ll inf = 0x3f3f3f3f3f3f3f3f;
#define debug cout << "i ak ioi" << "\n"
inline void print(int x) {if (x < 0) putchar('-'), x = -x; if(x > 9) print(x / 10); putchar(x % 10 + '0');}
inline char readchar() {static char buf[100000], *p1 = buf, *p2 = buf; return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2) ? EOF : *p1++;}
inline int read() { int res = 0, f = 0; char ch = gc();for (; !isdigit(ch); ch = gc()) f |= (ch == '-'); for (; isdigit(ch); ch = gc()) res = (res << 1) + (res << 3) + (ch ^ '0'); return f ? -res : res;}
#include <unordered_map>
int n, p, ans;
char ch; string t;
unordered_map <string, int> mp;
signed main()
{
n = read(), p = read();
for(int i = 1; i <= n; i++) {
string s; cin >> s; mp[s] = read();
}
while((ch = getchar()) && ch != EOF) {
if(!isdigit(ch) && !isalpha(ch)) ans = (ans + mp[t]) % p, t = "";
else t += ch;
}
printf("%d\n", ans);
return (0 - 0);
}
P1503 鬼子进村
用平衡树来维护被毁坏的房子，权值即为编号，用一个栈来取最后被毁坏的房子。
最后查询答案的时候就是 \(nxt - pre - 1\)。
/**
* author: TLE_Automation
* creater: 2022.9.7
**/
#include<cmath>
#include<queue>
#include<random>
#include<cstdio>
#include<bitset>
#include<cstring>
#include<iostream>
#include<algorithm>
#define gc getchar
using namespace std;
typedef long long ll;
const int N = 1e6 + 10;
const int mod = 998244353;
const ll inf = 0x3f3f3f3f3f3f3f3f;
#define debug cout << "i ak ioi" << "\n"
inline void print(int x) {if (x < 0) putchar('-'), x = -x; if(x > 9) print(x / 10); putchar(x % 10 + '0');}
inline char readchar() {static char buf[100000], *p1 = buf, *p2 = buf; return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2) ? EOF : *p1++;}
inline int read() { int res = 0, f = 0; char ch = gc();for (; !isdigit(ch); ch = gc()) f |= (ch == '-'); for (; isdigit(ch); ch = gc()) res = (res << 1) + (res << 3) + (ch ^ '0'); return f ? -res : res;}
int root, idx, x, y, z, res;
std::mt19937 rnd(114514);
namespace FHQ_Treap {
#define ls(x) tr[x].ls
#define rs(x) tr[x].rs
struct Node {int ls, rs, siz, val, key;} tr[N];
inline int newnode(int v) {int x = rnd(); tr[++idx] = (Node) {0, 0, 1, v, x}; return idx;}
inline void pushup(int p) {tr[p].siz = tr[ls(p)].siz + tr[rs(p)].siz + 1; }
inline void split(int p, int v, int &x, int &y) {
if(!p) {x = y = 0; return;}
if(tr[p].val <= v) x = p, split(rs(x), v, rs(x), y);
else y = p, split(ls(y), v, x, ls(y));
pushup(p);
}
inline int merge(int x, int y) {
if(!x || !y) return x + y;
if(tr[x].key < tr[y].key) { rs(x) = merge(rs(x), y), pushup(x); return x;}
else {ls(y) = merge(x, ls(y)), pushup(y); return y; }
}
inline void Insert(int v) {
split(root, v, x, y), z = newnode(v), root = merge(merge(x, z), y);
}
inline void del(int v) {
split(root, v, x, z), split(x, v - 1, x, y), y = merge(ls(y), rs(y)), root = merge(merge(x, y), z);
}
inline int getnum(int p, int k) {
if(tr[ls(p)].siz >= k) return getnum(ls(p), k);
if(tr[ls(p)].siz + 1 == k) return p;
return getnum(rs(p), k - tr[ls(p)].siz - 1);
}
inline int getpre(int v) {
split(root, v - 1, x, y), res = getnum(x, tr[x].siz), root = merge(x, y); return tr[res].val;
}
inline int getnxt(int v) {
split(root, v, x, y), res = getnum(y, 1), root = merge(x, y); return tr[res].val;
}
}
using namespace FHQ_Treap;
bitset <N> vis;
int stc[N], sc = 0;
signed main()
{
int n = read(), m = read();
Insert(0), Insert(n + 1);
for(int i = 1; i <= m; i++) {
char ch; cin >> ch;
if(ch == 'D') {
int x = read();
vis[x] = 1; stc[++sc] = x;
Insert(x);
}
else if(ch == 'R') {
vis[stc[sc]] = 0;
del(stc[sc--]);
}
else {
int x = read();
if(vis[x]) {puts("0"); continue;}
printf("%d\n", getnxt(x) - getpre(x) - 1);
}
}
return (0 - 0);
}
P3871 [TJOI2010]中位数
很明显的平衡树啊，随便敲敲就过了。
用一个 \(cnt\) 来维护当前平衡树中插入的数量，查询的时候分奇数偶数查询中位数即可。
/**
* author: TLE_Automation
* creater: 2022.9.7
**/
#include<cmath>
#include<queue>
#include<random>
#include<cstdio>
#include<bitset>
#include<cstring>
#include<iostream>
#include<algorithm>
#define gc getchar
using namespace std;
typedef long long ll;
const int N = 1e6 + 10;
const int mod = 998244353;
const ll inf = 0x3f3f3f3f3f3f3f3f;
#define debug cout << "i ak ioi" << "\n"
inline void print(int x) {if (x < 0) putchar('-'), x = -x; if(x > 9) print(x / 10); putchar(x % 10 + '0');}
inline char readchar() {static char buf[100000], *p1 = buf, *p2 = buf; return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2) ? EOF : *p1++;}
inline int read() { int res = 0, f = 0; char ch = gc();for (; !isdigit(ch); ch = gc()) f |= (ch == '-'); for (; isdigit(ch); ch = gc()) res = (res << 1) + (res << 3) + (ch ^ '0'); return f ? -res : res;}
int root, idx, x, y, z;
std::mt19937 rnd(114514);
namespace FHQ_Treap {
#define ls(x) tr[x].ls
#define rs(x) tr[x].rs
struct Node {int ls, rs, val, siz, key; } tr[N];
inline int newnode (int v) {int x = rnd(); tr[++idx] = (Node) {0, 0, v, 1, x}; return idx; }
inline void pushup (int p) {tr[p].siz = tr[ls(p)].siz + tr[rs(p)].siz + 1; }
inline void split(int p, int v, int &x, int &y) {
if(!p) {x = y = 0; return; }
if(tr[p].val <= v) x = p, split(rs(x), v, rs(x), y);
else y = p, split(ls(y), v, x, ls(y));
pushup(p);
}
inline int merge(int x, int y) {
if(!x || !y) return x + y;
if(tr[x].key < tr[y].key) { rs(x) = merge(rs(x), y); pushup(x); return x;}
else {ls(y) = merge(x, ls(y)); pushup(y); return y; }
}
inline void Insert(int v) {
split(root, v, x, y); z = newnode(v), root = merge(merge(x, z), y);
}
inline int getnum(int p, int k) {
if(tr[ls(p)].siz >= k) return getnum(ls(p), k);
else if(tr[ls(p)].siz + 1 == k) return p;
return getnum(rs(p), k - tr[ls(p)].siz - 1);
}
}
using namespace FHQ_Treap;
int cnt = 0;
signed main()
{
int n = read();
for(int i = 1; i <= n; i++) {
int o = read(); Insert(o);
cnt++;
}
int m = read();
for(int i = 1; i <= m; i++) {
string s; cin >> s;
if(s == "add") {
int o = read(); Insert(o);
cnt++;
}
else {
if(cnt & 1) printf("%d\n", tr[getnum(root, cnt / 2 + 1)].val);
else {
printf("%d\n", std::min(tr[getnum(root, cnt / 2)].val, tr[getnum(root, cnt / 2 + 1)].val));
}
}
}
return (0 - 0);
}
P2286 [HNOI2004]宠物收养场
平衡树模板，维护一个 cnt，记录当前是宠物树还是顾客树即可。
/**
* author: TLE_Automation
* creater: 2022.9.7
**/
#include<cmath>
#include<queue>
#include<random>
#include<cstdio>
#include<bitset>
#include<cstring>
#include<iostream>
#include<algorithm>
#define gc getchar
using namespace std;
typedef long long ll;
#define int long long
const int N = 1e6 + 10;
const int mod = 1000000;
const ll inf = 0x3f3f3f3f3f3f3f3f;
#define debug cout << "i ak ioi" << "\n"
inline void print(int x) {if (x < 0) putchar('-'), x = -x; if(x > 9) print(x / 10); putchar(x % 10 + '0');}
inline char readchar() {static char buf[100000], *p1 = buf, *p2 = buf; return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2) ? EOF : *p1++;}
inline int read() { int res = 0, f = 0; char ch = gc();for (; !isdigit(ch); ch = gc()) f |= (ch == '-'); for (; isdigit(ch); ch = gc()) res = (res << 1) + (res << 3) + (ch ^ '0'); return f ? -res : res;}
std::mt19937 rnd(114514);
int x, y, z, root, idx = 0, res;
namespace FHQ_Treap {
#define ls(x) tr[x].ls
#define rs(x) tr[x].rs
struct Node { int ls, rs, val, siz, key;} tr[N << 2];
inline int newnode(int v) { int x = rnd(); tr[++idx] = (Node) {0, 0, v, 1, x}; return idx;}
inline void pushup(int p) {tr[p].siz = tr[ls(p)].siz + tr[rs(p)].siz + 1;}
inline void split(int p, int v, int &x, int &y) {
if(!p) {x = y = 0; return;}
if(tr[p].val <= v) x = p, split(rs(x), v, rs(x), y);
else y = p, split(ls(y), v, x, ls(y));
pushup(p);
}
inline int merge(int x, int y) {
if(!x || !y) return x + y;
if(tr[x].key < tr[y].key) { rs(x) = merge(rs(x), y); pushup(x); return x;}
else {ls(y) = merge(x, ls(y)), pushup(y); return y; }
}
inline void Insert(int v) {
split(root, v, x, y); z = newnode(v), root = merge(merge(x, z), y);
}
inline void del(int v) {
split(root, v, x, z), split(x, v - 1, x, y), y = merge(ls(y), rs(y)); root = merge(merge(x, y), z);
}
inline int getnum(int p, int k) {
if(tr[ls(p)].siz >= k) return getnum(ls(p), k);
else if(tr[ls(p)].siz + 1 == k) return p;
else return getnum(rs(p), k - tr[ls(p)].siz - 1);
}
inline int getpre(int v) {
split(root, v - 1, x, y), res = getnum(x, tr[x].siz), root = merge(x, y); return tr[res].val;
}
inline int getnxt(int v) {
split(root, v, x, y), res = getnum(y, 1), root = merge(x, y); return tr[res].val;
}
}
using namespace FHQ_Treap;
int flag = 0;
ll ans;
signed main() {
int n = read();
Insert(0x3f3f3f3f), Insert(-0x3f3f3f3f);
for (int i = 1; i <= n; ++i) {
int a = read(), b = read();
if (flag == 0) Insert(b);
else if (flag < 0) {
if (!a) Insert(b);
else {
ll pick1 = getpre(b), pick2 = getnxt(b);
if (pick2 - b < b - pick1) del(pick2), (ans += pick2 - b) %= 1000000;
else del(pick1), (ans += b - pick1) %= 1000000;
}
} else if (flag > 0) {
if (a) Insert(b);
else {
ll pick1 = getpre(b), pick2 = getnxt(b);
if (pick2 - b < b - pick1) del(pick2), (ans += pick2 - b) %= 1000000;
else del(pick1), (ans += b - pick1) %= 1000000;
}
}
flag += 1 * (a ? 1 : -1);
}
printf("%lld\n", ans);
return 0;
}
P1486 [NOI2004] 郁闷的出纳员
用一个 \(tag\) 维护全局加的值，当插入一个数的时候，为了除去之前给的贡献，减去一个 tag，因为后面还会加上。
这里是分裂成 \(<x\) 和 \(>= x\) 的两颗子树。
/**
* author: TLE_Automation
* creater: 2022.9.7
**/
#include<cmath>
#include<queue>
#include<cstdio>
#include<random>
#include<bitset>
#include<cstring>
#include<iostream>
#include<algorithm>
#define gc getchar
using namespace std;
typedef long long ll;
const int N = 1e6 + 10;
const int mod = 998244353;
const ll inf = 0x3f3f3f3f3f3f3f3f;
#define debug cout << "i ak ioi" << "\n"
inline void print(int x) {if (x < 0) putchar('-'), x = -x; if(x > 9) print(x / 10); putchar(x % 10 + '0');}
inline char readchar() {static char buf[100000], *p1 = buf, *p2 = buf; return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2) ? EOF : *p1++;}
inline int read() { int res = 0, f = 0; char ch = gc();for (; !isdigit(ch); ch = gc()) f |= (ch == '-'); for (; isdigit(ch); ch = gc()) res = (res << 1) + (res << 3) + (ch ^ '0'); return f ? -res : res;}
int root, x, y, z, idx = 0, cnt = 0, n, Min, ans = 0;
std::mt19937 rnd(114514);
namespace FHQ_Treap {
#define ls(x) tr[x].ls
#define rs(x) tr[x].rs
struct Node {int ls, rs, val, siz, key; } tr[N];
inline int newnode(int v) {int x = rnd(); tr[++idx] = (Node) {0, 0, v, 1, x}; return idx;}
inline void pushup(int p) {tr[p].siz = tr[ls(p)].siz + tr[rs(p)].siz + 1; }
inline void split(int p, int v, int &x, int &y) {
if(!p) {x = y = 0; return ;}
if(tr[p].val < v) x = p, split(rs(x), v, rs(x), y);
else y = p, split(ls(y), v, x, ls(y));
pushup(p);
}
inline int merge(int x, int y) {
if(!x || !y) return x + y;
if(tr[x].key <tr[y].key) { rs(x) = merge(rs(x), y); pushup(x); return x;}
else {ls(y) = merge(x, ls(y)); pushup(y); return y;}
}
inline void Insert(int v) {
split(root, v, x, y), z = newnode(v), root = merge(merge(x, z), y);
}
inline void del(int v) {
split(root, v, x, z), split(x, v - 1, x, y), y = merge(ls(y), rs(y)); root = merge(merge(x, y), z);
}
inline int getnum(int p, int k) {
if(tr[ls(p)].siz >= k) return getnum(ls(p), k);
if(tr[ls(p)].siz + 1 == k) return p;
return getnum(rs(p), k - tr[ls(p)].siz - 1);
}
inline void leave() {
split(root, Min - cnt, x, y);
ans += tr[x].siz; root = y;
}
}
using namespace FHQ_Treap;
signed main()
{
n = read(), Min = read();
for(int i = 1; i <= n; i++) {
char ch; cin >> ch; int k = read();
if(ch == 'I') {
if(k >= Min) Insert(k - cnt);
}
else if(ch == 'A') cnt += k;
else if(ch == 'S') cnt -= k, leave();
else if(ch == 'F') printf("%d\n", k <= tr[root].siz ? tr[getnum(root, tr[root].siz - k + 1)].val + cnt : -1);
// cout << "##" << i << " " << cnt << " " << ans << "\n";
}
printf("%d\n", ans);
return (0 - 0);
}
P2343 宝石管理系统
插入和求 \(k\) 大， 纯纯的板子，太好写了。
/**
* author: TLE_Automation
* creater: 2022.9.7
**/
#include<cmath>
#include<queue>
#include<cstdio>
#include<random>
#include<bitset>
#include<cstring>
#include<iostream>
#include<algorithm>
#define gc getchar
using namespace std;
typedef long long ll;
const int N = 1e6 + 10;
const int mod = 998244353;
const ll inf = 0x3f3f3f3f3f3f3f3f;
#define debug cout << "i ak ioi" << "\n"
inline void print(int x) {if (x < 0) putchar('-'), x = -x; if(x > 9) print(x / 10); putchar(x % 10 + '0');}
inline char readchar() {static char buf[100000], *p1 = buf, *p2 = buf; return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2) ? EOF : *p1++;}
inline int read() { int res = 0, f = 0; char ch = gc();for (; !isdigit(ch); ch = gc()) f |= (ch == '-'); for (; isdigit(ch); ch = gc()) res = (res << 1) + (res << 3) + (ch ^ '0'); return f ? -res : res;}
std::mt19937 rnd(114514);
int root, x, y, z, idx = 0;
namespace FHQ_Treap {
#define ls(x) tr[x].ls
#define rs(x) tr[x].rs
struct Node {int ls, rs, val, siz, key; } tr[N];
inline int newnode(int v) {int x = rnd(); tr[++idx] = (Node) {0, 0, v, 1, x}; return idx; }
inline void pushup(int p) {tr[p].siz = tr[ls(p)].siz + tr[rs(p)].siz + 1; }
inline void split(int p, int v, int &x, int &y) {
if(!p) {x = y = 0; return; }
if(tr[p].val <= v) x = p, split(rs(x), v, rs(x), y);
else y = p, split(ls(y), v, x, ls(y));
pushup(p);
}
inline int merge(int x, int y) {
if(!x || !y) return x + y;
if(tr[x].key < tr[y].key) {rs(x) = merge(rs(x), y), pushup(x); return x;}
else {ls(y) = merge(x, ls(y)); pushup(y); return y; }
}
inline void Insert(int v) {
split(root, v, x, y), root = merge(merge(x, newnode(v)), y);
}
inline int getnum(int p, int k) {
if(tr[ls(p)].siz >= k) return getnum(ls(p), k);
if(tr[ls(p)].siz + 1 == k) return p;
return getnum(rs(p), k - tr[ls(p)].siz - 1);
}
}
using namespace FHQ_Treap;
signed main()
{
int n = read(), m = read();
for(int i = 1; i <= n; i++) x = read(), Insert(x);
while(m--) {
int opt = read(), k = read();
if(opt & 1) printf("%d\n", tr[getnum(root, tr[root].siz - k + 1)].val);
else Insert(k);
}
return (0 - 0);
}
