P2387 [NOI2014]魔法森林(LCT)

P2387 [NOI2014]魔法森林

LCT边权维护经典题

咋维护呢?边化为点,边权变点权。

本题中我们把边对关键字A进行排序,动态维护关键字B的最小生成树

加边后出现环咋办?

splay维护最大边的编号,找到最大边删除再加新边就ok辣

 

#include<cstdio>
#include<algorithm>
using namespace std;
inline void Swap(int &a,int &b){a^=b^=a^=b;}
inline int Min(int a,int b){return a<b?a:b;}
void read(int &x){
    static char c=getchar();x=0;
    while(c<'0'||c>'9') c=getchar();
    while('0'<=c&&c<='9') x=x*10+(c^48),c=getchar();
}
#define N 200005
struct edge{int f,t,A,B;}a[N];
int n,m,ans=2e9,ch[N][2],fa[N],val[N],s[N],rev[N],dad[N];
#define lc ch[x][0]
#define rc ch[x][1]
inline bool cmp(edge P,edge Q){return P.A<Q.A;}
int find(int x){return dad[x]==x?x:dad[x]=find(dad[x]);}
inline bool nrt(int x){return ch[fa[x]][0]==x||ch[fa[x]][1]==x;}
void up(int x){//维护平衡树上最大点权的点的编号(就是边化为的点)
    s[x]=x;
    if(val[s[lc]]>val[s[x]]) s[x]=s[lc];
    if(val[s[rc]]>val[s[x]]) s[x]=s[rc];    
}
void Rev(int x){Swap(lc,rc); rev[x]^=1;}
void down(int x){if(rev[x]) Rev(lc),Rev(rc),rev[x]=0;}
void Pre(int x){if(nrt(x))Pre(fa[x]); down(x);}
void turn(int x){
    int y=fa[x],z=fa[y],l=(ch[y][1]==x),r=l^1;
    if(nrt(y)) ch[z][ch[z][1]==y]=x;
    fa[ch[x][r]]=y; fa[y]=x; fa[x]=z;
    ch[y][l]=ch[x][r]; ch[x][r]=y;
    up(y); up(x);
}
void splay(int x){
    Pre(x);
    for(;nrt(x);turn(x)){
        int y=fa[x],z=fa[y];
        if(nrt(y)) turn(((ch[z][1]==y)^(ch[y][1]==x))?x:y);
    }
}
void access(int x){for(int y=0;x;y=x,x=fa[x]) splay(x),rc=y,up(x);}
void makert(int x){access(x);splay(x);Rev(x);}
int findrt(int x){
    access(x);splay(x);down(x);
    while(lc) x=lc,down(x);
    splay(x); return x;
}
void link(int x,int y){makert(x); if(findrt(y)!=x)fa[x]=y;}
void cut(int x,int y){
    makert(x);
    if(findrt(y)==x&&fa[y]==x&&!ch[y][0]) fa[y]=rc=0,up(x);
}
void split(int x,int y){makert(x);access(y);splay(y);}
int main(){
    read(n);read(m); register int i;
    for(i=1;i<=n+m;++i) dad[i]=i;
    for(i=1;i<=m;++i)
        read(a[i].f),read(a[i].t),read(a[i].A),read(a[i].B);
    sort(a+1,a+m+1,cmp);
    for(i=1;i<=m;++i) val[n+i]=a[i].B;
    for(i=1;i<=m;++i){
        int r1=find(a[i].f),r2=find(a[i].t),e;//并查集维护是否连通
        if(r1==r2){
            split(a[i].f,a[i].t); e=s[a[i].t];
            if(val[e]<=a[i].B) continue;
            cut(a[e-n].f,e); cut(e,a[e-n].t);//删掉最大边
        }else dad[r1]=r2;
        link(a[i].f,i+n); link(i+n,a[i].t);
        if(find(1)==find(n)) split(n,1),ans=Min(ans,a[i].A+val[s[1]]);
    }
    if(ans>=2e9) puts("-1");
    else printf("%d",ans);
    return 0;
}

 

posted @ 2019-03-21 19:46  kafuuchino  阅读(170)  评论(0编辑  收藏  举报