[题解] bzoj 3674 可持久化并查集加强版
- 传送门 -
http://www.lydsy.com/JudgeOnline/problem.php?id=3674
3674: 可持久化并查集加强版
Time Limit: 15 Sec Memory Limit: 256 MB
Submit: 4218 Solved: 1557
Description
自从zkysb出了可持久化并查集后……
hzwer:乱写能AC，暴力踩标程
KuribohG：我不路径压缩就过了！
ndsf：暴力就可以轻松虐！
zky:……
n个集合 m个操作
操作：
1 a b 合并a,b所在集合
2 k 回到第k次操作之后的状态(查询算作操作)
3 a b 询问a,b是否属于同一集合，是则输出1否则输出0
请注意本题采用强制在线,所给的a,b,k均经过加密,加密方法为x = x xor lastans,lastans的初始值为0
0<n,m<=2*10^5
Sample Input
5 6
1 1 2
3 1 2
2 1
3 0 3
2 1
3 1 2
Sample Output
1
0
1
- 思路 -
未加强的在3673,但是啥优化都不加也能过.
然后做到这题我才发现并查集有种叫按秩合并的优化...我是傻子吗...
看了第一篇讲解说就是把rank小的接在大的下面
但是rank是什么....
于是我就把编号小的接在大的下面...然后就过了...
可是这和我看到的AC代码好像不一样...
然后又看了一篇讲解,发现按秩合并是指把数量小的树接到大的上去,通常是用高度来取代这个数量
...
哦我终于看懂了标程了...
再打一遍,然后发现慢了一倍...
哦仔细想想按编号来优化也是可以的,除非是8,7,6,5,4,3,2,1这样倒着合并会合出一条链,但是大部分情况下还是可行的,而且也不需要维护什么东西(大概就是这里节省了时间).
路径压缩...不存在的...怎么可能说我懒得打
细节见代码.
- 代码 -
按编号优化:
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <vector>
#include <ctime>
#include <cstdlib>
#define pii pair<int, int>
#define mp make_pair
#define ft first
#define sd second
#define ls C[rt][0]
#define rs C[rt][1]
using namespace std;
template <typename ty> void read(ty &x) {
x = 0; int f = 1; char ch = getchar();
while (ch > '9' || ch < '0') { if (ch == '-') f = -1; ch = getchar(); }
while (ch >= '0' && ch <= '9') { x = x*10 + ch - '0'; ch = getchar(); }
x *= f;
}
template <typename ty> ty Max(ty a, ty b) { return a > b ? a : b; }
template <typename ty> ty Min(ty a, ty b) { return a < b ? a : b; }
template <typename ty> int Chkmin(ty &a, ty b) { return a > b ? a = b, 1 : 0; }
template <typename ty> int Chkmax(ty &a, ty b) { return a < b ? a = b, 1 : 0; }
typedef long long LL;
typedef double db;
const int inf = 0x7fffffff;
const int N = 2e5 + 16;
int F[N], V[N*20], C[N*20][2], RT[N], P[N];
int n, m, sz, lastans;
void build(int &rt, int l, int r) {
rt = ++sz;
if (l == r) { V[rt] = l; return; }
int mid = l + r >> 1;
build(C[rt][0], l, mid);
build(C[rt][1], mid + 1, r);
}
int find(int rt) { return F[rt] == rt ? rt : F[rt] = find(F[rt]); }
int query(int rt, int l, int r, int pos) {
if (l == r) return V[rt];
int mid = l + r >> 1;
if (pos <= mid) return query(C[rt][0], l, mid, pos);
else return query(C[rt][1], mid + 1, r, pos);
}
void change(int &now, int pre, int l, int r, int pos, int v) {
now = ++sz;
C[now][0] = C[pre][0]; C[now][1] = C[pre][1];
if (l == r) { V[now] = v; return; }
int mid = l + r >> 1;
if (pos <= mid) change(C[now][0], C[pre][0], l, mid, pos, v);
else change(C[now][1], C[pre][1], mid + 1, r, pos, v);
}
int get(int rt, int a) {
int tmp = a;
while (true) {
tmp = query(RT[rt], 1, n, tmp);
if (query(RT[rt], 1, n, tmp) != tmp)
a = tmp;
else return tmp;
}
}
int main () {
read(n); read(m);
build(RT[0], 1, n);
for (int i = 1; i <= m; ++ i) {
F[i] = i;
int opt, a, b, k;
read(opt);
if (opt == 1) {
read(a); read(b);
a ^= lastans, b ^= lastans;
int pre = find(i - 1);
int tmp1, tmp2;
tmp1 = get(pre, a);
tmp2 = get(pre, b);
if (tmp1 == tmp2) { F[i] = i - 1; continue; }
if (tmp1 < tmp2) swap(tmp1, tmp2);
change(RT[i], RT[pre], 1, n, tmp1, tmp2);
}
if (opt == 2) {
read(k);
k ^= lastans;
F[i] = find(k);
}
if (opt == 3) {
F[i] = find(i - 1);
read(a); read(b);
a ^= lastans, b ^= lastans;
int tmp, tmp1, tmp2;
tmp1 = get(F[i], a);
tmp2 = get(F[i], b);
if (tmp1 == tmp2) lastans = 1, printf("1\n");
else lastans = 0, printf("0\n");
}
}
return 0;
}
按秩合并:
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <vector>
#include <ctime>
#include <cstdlib>
#define pii pair<int, int>
#define mp make_pair
#define ft first
#define sd second
#define ls C[rt][0]
#define rs C[rt][1]
using namespace std;
template <typename ty> void read(ty &x) {
x = 0; int f = 1; char ch = getchar();
while (ch > '9' || ch < '0') { if (ch == '-') f = -1; ch = getchar(); }
while (ch >= '0' && ch <= '9') { x = x*10 + ch - '0'; ch = getchar(); }
x *= f;
}
template <typename ty> ty Max(ty a, ty b) { return a > b ? a : b; }
template <typename ty> ty Min(ty a, ty b) { return a < b ? a : b; }
template <typename ty> int Chkmin(ty &a, ty b) { return a > b ? a = b, 1 : 0; }
template <typename ty> int Chkmax(ty &a, ty b) { return a < b ? a = b, 1 : 0; }
typedef long long LL;
typedef double db;
const int inf = 0x7fffffff;
const int N = 2e5 + 16;
int F[N], V[N*20], C[N*20][2], D[N*20], RT[N], P[N];
int n, m, sz, lastans;
void build(int &rt, int l, int r) {
rt = ++sz;
if (l == r) { V[rt] = l; return; }
int mid = l + r >> 1;
build(C[rt][0], l, mid);
build(C[rt][1], mid + 1, r);
}
int find(int rt) { return F[rt] == rt ? rt : F[rt] = find(F[rt]); }
int query(int rt, int l, int r, int pos) {
if (l == r) return V[rt];
int mid = l + r >> 1;
if (pos <= mid) return query(C[rt][0], l, mid, pos);
else return query(C[rt][1], mid + 1, r, pos);
}
void change(int &now, int pre, int l, int r, int pos, int v) {
now = ++sz;
C[now][0] = C[pre][0]; C[now][1] = C[pre][1];
if (l == r) { V[now] = v; return; }
int mid = l + r >> 1;
if (pos <= mid) change(C[now][0], C[pre][0], l, mid, pos, v);
else change(C[now][1], C[pre][1], mid + 1, r, pos, v);
}
int get(int rt, int a) {
int tmp = a;
while (true) {
tmp = query(RT[rt], 1, n, tmp);
if (query(RT[rt], 1, n, tmp) != tmp)
a = tmp;
else return tmp;
}
}
void add(int rt, int l, int r, int pos) {
if (l == r) { D[rt] ++; return; }
int mid = l + r >> 1;
if (pos <= mid) add(C[rt][0], l, mid, pos);
else add(C[rt][1], mid + 1, r, pos);
}
int main () {
read(n); read(m);
build(RT[0], 1, n);
for (int i = 1; i <= m; ++ i) {
F[i] = i;
int opt, a, b, k;
read(opt);
if (opt == 1) {
read(a); read(b);
a ^= lastans, b ^= lastans;
int pre = find(i - 1);
int tmp1, tmp2;
tmp1 = get(pre, a);
tmp2 = get(pre, b);
if (tmp1 == tmp2) { F[i] = i - 1; continue; }
if (D[tmp1] > D[tmp2]) swap(tmp1, tmp2);
change(RT[i], RT[pre], 1, n, tmp1, tmp2);
if (D[tmp1] == D[tmp2]) add(RT[i], 1, n, tmp2);
}
if (opt == 2) {
read(k);
k ^= lastans;
F[i] = find(k);
}
if (opt == 3) {
F[i] = find(i - 1);
read(a); read(b);
a ^= lastans, b ^= lastans;
int tmp, tmp1, tmp2;
tmp1 = get(F[i], a);
tmp2 = get(F[i], b);
if (tmp1 == tmp2) lastans = 1, printf("1\n");
else lastans = 0, printf("0\n");
}
}
return 0;
}
