BZOJ 3514: Codechef MARCH14 GERALD07加强版( LCT + 主席树 )

从左到右加边, 假如+的边e形成环, 那么记下这个环上最早加入的边_e, 当且仅当询问区间的左端点> _e加入的时间, e对答案有贡献(脑补一下). 然后一开始是N个连通块, 假如有x条边有贡献, 答案就是N-x. 用LCT维护加边, 可持久化线段树维护询问. O(NlogN)

------------------------------------------------------------------------------------

#include<cstdio>

#include<cctype>

#include<cstring>

#include<algorithm>

 

using namespace std;

 

const int maxn = 200009;

 

int N, M, Q, typ;

int ntr[maxn], par[maxn], sm[maxn], buf[16];

 

inline int getint() {

char c = getchar();

for(; !isdigit(c); c = getchar());

int ret = 0;

for(; isdigit(c); c = getchar())

ret = ret * 10 + c - '0';

return ret;

}

 

inline void putint(int x) {

if(!x) {

puts("0"); return;

}

int n = 0;

for(; x; x /= 10)

buf[n++] = x % 10;

while(n--)

putchar(buf[n] + '0');

puts("");

}

 

int Find(int x) {

return x == par[x] ? x : par[x] = Find(par[x]);

}

 

struct edge {

int u, v;

} e[maxn];

 

struct Node* Null;

 

struct Node {

Node *ch[2], *p, *fa, *mn;

bool iR, rev;

int n;

inline void setc(Node* c, int t) {

(ch[t] = c)->p = this;

}

inline void pd() {

if(rev) {

swap(ch[0], ch[1]);

ch[0]->rev ^= 1;

ch[1]->rev ^= 1;

rev = false;

}

}

inline void upd() {

mn = this;

if(ch[0] != Null && mn->n > ch[0]->mn->n) mn = ch[0]->mn;

if(ch[1] != Null && mn->n > ch[1]->mn->n) mn = ch[1]->mn;

}

inline int d() {

return this == p->ch[1];

}

inline void setR() {

fa = p, p = Null;

iR = true;

}

} pool[maxn << 1], *T, *V[maxn], *stk[maxn << 1];

 

void Init_LCT() {

T = pool;

T->ch[0] = T->ch[1] = T;

T->fa = T->p = T->mn = T;

T->n = maxn;

Null = T++;

}

 

Node* Newnode(int n) {

T->n = n;

T->ch[0] = T->ch[1] = Null;

T->fa = T->p = Null;

T->iR = true, T->rev = false;

return T++;

}

 

void Rot(Node* t) {

Node* p = t->p;

p->pd(), t->pd();

int d = t->d();

p->p->setc(t, p->d());

p->setc(t->ch[d ^ 1], d);

t->setc(p, d ^ 1);

p->upd();

if(p->iR) {

p->iR = false;

t->iR = true;

t->fa = p->fa;

}

}

 

void Splay(Node* t) {

int n = 0;

for(Node* o = t; o != Null; o = o->p) stk[n++] = o;

while(n--) stk[n]->pd();

for(Node* p = t->p; p != Null; p = t->p) {

if(p->p != Null)

p->d() != t->d() ? Rot(t) : Rot(p);

Rot(t);

}

t->upd();

}

 

void Access(Node* t) {

for(Node* o = Null; t != Null; o = t, t = t->fa) {

Splay(t);

t->ch[1]->setR();

t->setc(o, 1);

}

}

 

void makeRoot(Node* t) {

Access(t);

Splay(t);

t->rev ^= 1;

}

 

Node* Path(Node* u, Node* v) {

makeRoot(u);

Access(v), Splay(v);

return v;

}

 

void Join(Node* u, Node* v) {

makeRoot(u);

u->fa = v;

}

 

void Cut(Node* u, Node* v) {

makeRoot(u);

Access(v), Splay(v);

u->p = Null, u->setR();

v->setc(Null, 0), v->upd();

}

 

namespace F {

int Val;

struct Node {

Node *lc, *rc;

int n;

} pool[maxn * 80], *pt, *Null, *Root[maxn];

void Init() {

pt = pool;

pt->lc = pt->rc = pt;

pt->n = 0;

Root[0] = Null = pt++;

}

Node* Modify(Node* t, int l, int r) {

Node* o = pt++;

o->n = t->n + 1;

if(l != r) {

int m = (l + r) >> 1;

if(Val <= m) {

o->lc = Modify(t->lc, l, m);

o->rc = t->rc;

} else {

o->lc = t->lc;

o->rc = Modify(t->rc, m + 1, r);

}

}

return o;

}

}

 

int main() {

N = getint(), M = getint();

Q = getint(), typ = getint();

Init_LCT();

for(int i = 0; i < N; i++)

par[i] = i, V[i] = Newnode(maxn);

for(int i = 1; i <= M; i++) {

int &u = e[i].u, &v = e[i].v;

u = getint() - 1, v = getint() - 1;

if(u == v) {

ntr[i] = maxn;

continue;

}

int _u = Find(u), _v = Find(v);

Node* t = Newnode(i);

if(_u == _v) {

Node* mn = Path(V[u], V[v])->mn;

Cut(V[e[mn->n].u], mn);

Cut(V[e[mn->n].v], mn);

ntr[i] = mn->n;

} else {

ntr[i] = 0;

par[_u] = _v;

}

Join(t, V[u]), Join(t, V[v]);

}

F::Init();

for(int i = 1; i <= M; i++) if(ntr[i] >= 1 && ntr[i] <= M) {

F::Val = ntr[i];

F::Root[i] = F::Modify(F::Root[i - 1], 1, M);

} else 

F::Root[i] = F::Root[i - 1];

sm[0] = 0;

for(int i = 1; i <= M; i++)

sm[i] = sm[i - 1] + !ntr[i];

int ans = 0, l, r;

while(Q--) {

l = getint(), r = getint();

if(typ)

l ^= ans, r ^= ans;

int p = l - 1;

ans = sm[r] - sm[p];

F::Node *L = F::Root[p], *R = F::Root[r];

l = 1, r = M;

while(l < r) {

int m = (l + r) >> 1;

if(m <= p) {

ans += R->lc->n - L->lc->n;

L = L->rc, R = R->rc;

l = m + 1;

} else {

L = L->lc, R = R->lc;

r = m;

}

}

putint(ans = N - ans);

}

return 0;

}

------------------------------------------------------------------------------------

3514: Codechef MARCH14 GERALD07加强版

Time Limit: 40 Sec  Memory Limit: 256 MB
Submit: 685  Solved: 232
[Submit][Status][Discuss]

Description

N个点M条边的无向图，询问保留图中编号在[l,r]的边的时候图中的联通块个数。

Input

第一行四个整数N、M、K、type，代表点数、边数、询问数以及询问是否加密。
接下来M行，代表图中的每条边。
接下来K行，每行两个整数L、R代表一组询问。对于type=0的测试点，读入的L和R即为询问的L、R；对于type=1的测试点，每组询问的L、R应为L xor lastans和R xor lastans。

Output

 K行每行一个整数代表该组询问的联通块个数。

Sample Input

3 5 4 0
1 3
1 2
2 1
3 2
2 2
2 3
1 5
5 5
1 2

Sample Output

2
1
3
1

HINT

对于100%的数据，1≤N、M、K≤200,000。

Source

By zhonghaoxi