【BZOJ 2594】【WC 2006】水管局长数据加强版
离线后倒过来做,这样就跟魔法森林差不多了,缩边为点就可以统计边的权值了。
1A真是爽,可惜常数炸上了天,这是滥用stl容器和无脑link,cut的后果
#include<map> #include<cstdio> #include<cstring> #include<algorithm> using namespace std; const int N = 100003; const int M = 1000003; void read(int &k) { k = 0; int fh = 1; char c = getchar(); for(; c < '0' || c > '9'; c = getchar()) if (c == '-') fh = -1; for(; c >= '0' && c <= '9'; c = getchar()) k = (k << 1) + (k << 3) + c - '0'; k = k * fh; } struct node *null; struct node { node *ch[2], *fa; int mx, mxid, d, id; short rev; bool pl() {return fa->ch[1] == this;} bool check() {return fa == null || (fa->ch[0] != this && fa->ch[1] != this);} void push() {if (rev) {rev = 0; swap(ch[0], ch[1]); ch[0]->rev ^= 1; ch[1]->rev ^= 1;}} void count() { mxid = id; mx = d; if (ch[0]->mx > mx) {mx = ch[0]->mx; mxid = ch[0]->mxid;} if (ch[1]->mx > mx) {mx = ch[1]->mx; mxid = ch[1]->mxid;} } void setc(node *r, bool c) {ch[c] = r; r->fa = this;} } pool[N + M]; struct Quee{int op, u, v;} Q[N]; struct Data{int u, v, e;} E[M]; struct maa{ int u, v; bool operator < (const maa data) const {return u == data.u ? v < data.v : u < data.u;} }; bool cmp(Data A, Data B) {return A.e < B.e;} map <maa, int> ma; bool cannot[M]; namespace LCT { int fa[N]; int find(int x) {return fa[x] == x ? x : (fa[x] = find(fa[x]));} void rotate(node *r) { node *f = r->fa; bool c = r->pl(); if (f->check()) r->fa = f->fa; else f->fa->setc(r, f->pl()); f->setc(r->ch[!c], c); r->setc(f, !c); f->count(); } void update(node *r) {if (!r->check()) update(r->fa); r->push();} void splay(node *r) { update(r); for(; !r->check(); rotate(r)) if (!r->fa->check()) rotate(r->pl() == r->fa->pl() ? r->fa : r); r->count(); } node *access(node *r) {node *y = null; for(; r != null; y = r, r = r->fa) splay(r), r->ch[1] = y; return y;} void changert(node *r) {access(r)->rev ^= 1; splay(r);} void link(node *r, node *t) {changert(r); r->fa = t;} void cut(node *r, node *t) {changert(r); access(t); splay(t); t->ch[0]->fa = null; t->ch[0] = null;} node *findrt(node *r) {access(r); splay(r); while (r->ch[0] != null) r = r->ch[0]; return r;} void newnode(int k, int num) { pool[k].ch[0] = pool[k].ch[1] = pool[k].fa = null; pool[k].id = pool[k].mxid = k; pool[k].d = pool[k].mx = num; pool[k].rev = 0; } int A[N]; void Build(int n, int m, int q) { null = &pool[0]; null->ch[0] = null->ch[1] = null->fa = null; null->id = null->d = null->mx = null->mxid = null->rev = 0; int u, v, e; for(int i = 1; i <= n; ++i) newnode(i, 0); for(int i = 1; i <= m; ++i) { read(u); read(v); read(e); if (u > v) swap(u, v); E[i] = (Data) {u, v, e}; } sort(E + 1, E + m + 1, cmp); for(int i = 1; i <= m; ++i) { ma[(maa) {E[i].u, E[i].v}] = i; newnode(i + n, E[i].e); } for(int i = 1; i <= q; ++i) { read(Q[i].op); read(Q[i].u); read(Q[i].v); if (Q[i].u > Q[i].v) swap(Q[i].u, Q[i].v); if (Q[i].op == 2) cannot[ma[(maa) {Q[i].u, Q[i].v}]] = 1; } node *tox, *toy, *toe; int cnt = 0, x, y, fx, fy; for(int i = 1; i <= n; ++i) fa[i] = i; for(int i = 1; i <= m; ++i) { x = E[i].u; y = E[i].v; fx = find(x); fy = find(y); if (fx != fy && !cannot[e = ma[(maa) {x, y}]]) { ++cnt; fa[fx] = fy; tox = &pool[x]; toy = &pool[y]; toe = &pool[e + n]; link(tox, toe); link(toy, toe); if (cnt == n - 1) break; } } cnt = 0; for(int i = q; i >= 1; --i) { if (Q[i].op == 1) { tox = &pool[Q[i].u]; toy = &pool[Q[i].v]; changert(tox); access(toy); splay(toy); A[++cnt] = toy->mx; } else { tox = &pool[Q[i].u]; toy = &pool[Q[i].v]; e = ma[(maa) {Q[i].u, Q[i].v}]; changert(tox); access(toy); splay(toy); if (E[e].e < toy->mx) { u = toy->mxid - n; tox = &pool[E[u].u]; toy = &pool[E[u].v]; toe = &pool[u + n]; cut(tox, toe); cut(toy, toe); tox = &pool[Q[i].u]; toy = &pool[Q[i].v]; toe = &pool[e + n]; link(tox, toe); link(toy, toe); } } } for(int i = cnt; i; --i) printf("%d\n", A[i]); } } int n, m, q; int main() { read(n); read(m); read(q); LCT::Build(n, m, q); return 0; }
模板一定不能写残啊,想起一个月前Round1Day1调LCT的模板调了4h,查出各种手残的错误QAQ然而并没有什么用。
NOI 2017 Bless All