$LCT+枚举$ 复习一下$LCT$模板。
先以$Ai$为关键字$sort$,然后$Ai$从小到大枚举每条边,看能否构成环,构不成则加边,构成则判断,判断过了就切断$Bi$最大的边。
我的边是编号为$i+n$的点,忘了这点调了好久$QAQ$ $sosad$
#include<cstdio>
#include<cstring>
#include<algorithm>
#define N 150003
#define read(x) x=getint()
using namespace std;
inline int getint() {int k = 0, 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 * 10 + c - ‘0‘; return k * fh;}
struct nodeE {int a, b, x, y;} E[N];
struct node *null;
struct node {
node *ch[2], *fa;
int d, pos;
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() {
pos = d;
if (E[ch[0]->pos].b > E[pos].b) pos = ch[0]->pos;
if (E[ch[1]->pos].b > E[pos].b) pos = ch[1]->pos;
}
void setc(node *r, bool c) {ch[c] = r; r->fa = this;}
} *rt[N];
node pool[N];
int n, m, tot = 0;
namespace LCT {
int ans = 0x7fffffff;
bool cmp(nodeE X, nodeE Y) {return X.a < Y.a;}
node *newnode(int num = 0) {
node *t = &pool[++tot];
t->ch[0] = t->ch[1] = t->fa = null;
t->d = t->pos = num; t->rev = 0;
return t;
}
void Build() {
null = &pool[0];
null->ch[0] = null->ch[1] = null->fa = null;
null->d = null->pos = null->rev = 0;
read(n); read(m);
for(int i = 1; i <= m; ++i)
{read(E[i].x); read(E[i].y); read(E[i].a); read(E[i].b);}
sort(E + 1, E + m + 1, cmp);
for(int i = 1; i <= n; ++i)
rt[i] = newnode();
for(int i = 1; i <= m; ++i)
rt[n + i] = newnode(i);
E[0].b = 0;
}
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;}
int ask(node *r, node *t) {changert(r); access(t); splay(t); return t->pos;}
void work(int u, int v, int edge) {
if (findrt(rt[u]) == findrt(rt[v])) {
int k = ask(rt[u], rt[v]);
if (E[k].b > E[edge - n].b) {
cut(rt[u], rt[k + n]);
cut(rt[v], rt[k + n]);
link(rt[u], rt[edge]);
link(rt[v], rt[edge]);
}
} else {
link(rt[u], rt[edge]);
link(rt[v], rt[edge]);
}
if (findrt(rt[1]) == findrt(rt[n]))
ans = min(ans, E[edge - n].a + E[ask(rt[1], rt[n])].b);
}
void AC() {printf("%d\n", ans == 0x7fffffff ? -1 : ans);}
}
int main() {
LCT::Build();
for(int i = 1; i <= m ;++i)
LCT::work(E[i].x, E[i].y, i + n);
LCT::AC();
return 0;
}
我的代码就是一堵墙,让$300$行的$LinkCutTree$在压行大法前颤抖吧~~~
【BZOJ 3669】【NOI 2014】魔法森林 LCT+枚举边
原文:http://www.cnblogs.com/abclzr/p/5350445.html