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NOI 2014魔法森林

2015-01-26 17:32 309 查看
解法大致同BZOJ2594

首先先将a排序,逐边加入,每次维护一条链上最大的b,再次加入时,如果没有连通则直接链上,否则必定会生成环,那么就将b最大的删去即可

#include <cstdio>
#include <algorithm>

using namespace std;

const int MAXN = 50001;
const int MAXM = 100001;
const int INF = 1000001;

void READ(int &x)
{
x = 0;
char c;
do c = getchar();while (c > '9' || c < '0');
do x = x*10+c-48, c = getchar(); while (c <= '9' && c >= '0');
};

struct Edge
{
int u,v,a,b;
};

Edge Edges[MAXM];

int n,m;
int f[MAXN];

bool cmpa(Edge A,Edge B)
{
return A.a < B.a;
}

struct Node *null;
struct Node
{
int id;
Node *ch[2],*f;
int vA,vB;
int maxA,maxB,maxBlocal;
int size;
bool rev;
void maintain()
{
size = 1+ch[0]->size+ch[1]->size;
maxB = vB, maxBlocal = id;
if (ch[0]->maxB > maxB) maxB = ch[0]->maxB, maxBlocal = ch[0]->maxBlocal;
if (ch[1]->maxB > maxB) maxB = ch[1]->maxB, maxBlocal = ch[1]->maxBlocal;
maxA = max(vA,ch[0]->maxA), maxA = max(maxA,ch[1]->maxA);
}
void pushdown()
{
if (rev)
{
rev ^= 1;
ch[0]->rev ^= 1, ch[1]->rev ^= 1;
swap(ch[0],ch[1]);
}
}
void setc(Node *c,int d){ch[d] = c, c->f = this;}
bool check(){return f != null && (f->ch[0] == this || f->ch[1] == this);}
int d(){return f->ch[1] == this ? 1 : 0;}
Node (int _id,int _vA,int _vB):id(_id),vA(_vA),vB(_vB){ch[0] = ch[1] = f = null, rev = false, maintain();}
Node (){ch[0] = ch[1] = NULL, rev = false, size = 0, maxA = maxB = -INF;}
};

Node *p[MAXN+MAXM];

void rotate(Node *p)
{
Node *x = p->f;
x->pushdown();p->pushdown();
int d = p->d();
if (!x->check()) p->f = x->f;
else x->f->setc(p,x->d());
x->setc(p->ch[d^1],d);
p->setc(x,d^1);
x->maintain();
}

void Splay(Node *p)
{
p->pushdown();
while (p->check())
{
Node *x = p->f;
if (!x->check()) rotate(p);
else x->d() == p->d() ? (rotate(x),rotate(p)) : (rotate(p),rotate(p));
}
p->maintain();
}

void Access(Node *p)
{
for (Node *q = null; p != null; p = p->f)
{
Splay(p);
p->setc(q,1);
(q = p)->maintain();
}
}

inline void MakeRoot(Node *p)
{
Access(p);Splay(p);p->rev ^= 1;
}

void Link(Node *p,Node *q)
{
MakeRoot(p);p->f = q;
}

void Cut(Node *p,Node *q)
{
MakeRoot(p);
Access(q);Splay(q);
p->f = q->ch[0] = null;
q->maintain();
}

Node* Query(Node *p,Node *q)
{
MakeRoot(p);
Access(q);Splay(q);
return q;
}

int find(int x)
{
return x == f[x] ? f[x] : f[x] = find(f[x]);
}

int main()
{
null = new Node();
READ(n);READ(m);
for (int i=1;i<=m;i++) READ(Edges[i].u),READ(Edges[i].v),READ(Edges[i].a),READ(Edges[i].b);
sort(Edges+1,Edges+m+1,cmpa);
for (int i=1;i<=n;i++) f[i] = i;
for (int i=1;i<=n+m;i++) p[i] = new Node(i,0,0);
int ans = INF;
for (int i=1;i<=m;i++)
{
int Su = find(Edges[i].u);
int Sv = find(Edges[i].v);
Node *temp;
if (Su != Sv)
{
f[Su] = Sv;
p[i+n]->vA = Edges[i].a, p[i+n]->vB = Edges[i].b;
Link(p[Edges[i].u],p[i+n]);Link(p[Edges[i].v],p[i+n]);
}
else
{
temp = Query(p[Edges[i].u],p[Edges[i].v]);
if (Edges[i].b < temp->maxB)
{
int point = temp->maxBlocal-n;
Cut(p[point+n],p[Edges[point].u]);Cut(p[point+n],p[Edges[point].v]);
p[i+n] = new Node(i+n,Edges[i].a,Edges[i].b);
Link(p[i+n],p[Edges[i].u]);Link(p[i+n],p[Edges[i].v]);
}
}
if (find(1) == find(n)) temp = Query(p[1],p
), ans = min(ans,temp->maxA+temp->maxB);
}
if (find(1) != find(n))
{
puts("-1");
return 0;
}
printf("%d\n",ans);
return 0;
}
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