BZOJ1787: [Ahoi2008]Meet 紧急集合【LCA】

dzy posted @ 2013年6月11日 21:43 in BZOJ with tags LCA bzoj 倍增 , 2488 阅读

给你一颗无根树。每条边的路径长是1。

每次询问给你三个点。要求你找到一个点。使其它三个点到这个点的路径和最小。

看到这种题就很快联想到LCA了。然后看看样例然后随便手造几组数据,发现求的点一定在其中两个点的LCA。

那么就好了。每次求出3个LCA。然后枚举。算出距离和。

介于倍增比rmq好写的多。就没写rmq。不过倍增的话每次求lca是logn的。每次询问差不多要做6次LCA。那么6logn。

如果是rmq的话每次询问时O(1)。预处理都是nlogn。那么常数一下就跪了。刷rank1也没戏了、不过还有个rank2= =还是玩不过seter、、跪烂、、

 





#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
char buf[21000000],*pt=buf,*o=buf;
int p[1000005],n1[1000005],h[500005]={0},q[500005],d[500005],v[500005]={0};
int lg[500005],a[500005][20],n,m,tot=1;
inline int getint(){    int r=0; while(*pt<48||*pt>57)pt++;
    while(*pt>47&&*pt<58)r=r*10+*pt++-48; return r;
}inline void print(int x){
    char str[11],*p=str; if(!x)*o++=48;
    else{ while(x) *p++=x%10+48,x/=10; while(p--!=str)*o++=*p;}
}inline void ae(int a,int b){
    p[tot]=b; n1[tot]=h[a]; h[a]=tot++; p[tot]=a; n1[tot]=h[b]; h[b]=tot++;
}inline void pre(){
    int s=0,e=1; q[0]=1; d[1]=1; v[1]=1;
    while(s<e){
        for(int i=h[q[s++]];i;i=n1[i]) if(!v[p[i]])
            v[p[i]]=1,q[e++]=p[i],d[p[i]]=d[q[s-1]]+1,a[p[i]][0]=q[s-1];
    }for(int j=1;j<20;j++) for(int i=1;i<=n;i++) a[i][j]=a[a[i][j-1]][j-1];
}inline int lca(int u,int v){
    if(d[u]>d[v]){ u^=v; v^=u; u^=v;}
    int k=d[v]-d[u],j=0; while(k){ if(k&1)v=a[v][j]; j++; k>>=1;}
    if(u==v)return u;
    for(int i=lg[d[u]-1];~i;i--)
        if(a[u][i]!=a[v][i]) u=a[u][i],v=a[v][i];
    return a[u][0]; 
}int main(){
    fread(pt,1,21000000,stdin);  
    n=getint();m=getint();
    for(int i=0;(1<<i)<=n;i++)
        for(int j=(1<<i);j<(1<<i+1)&&j<=n;j++)lg[j]=i;
    int u,v,w,a1,a2,a3,a4,a5,An1,An2;
    for(int i=1;i<n;i++) u=getint(),v=getint(),ae(u,v); 
    pre();
    for(int i=1;i<=m;i++){
        u=getint(),v=getint(),w=getint();
        a1=lca(u,v),a2=lca(u,w),a3=lca(v,w),a4=d[u]+d[v]+d[w];
        An1=a1,An2=a4-d[a1]-(d[lca(w,a1)]<<1);
        if((a5=a4-d[a2]-(d[lca(v,a2)]<<1))<An2) An2=a5,An1=a2;
        if((a5=a4-d[a3]-(d[lca(u,a3)]<<1))<An2) An2=a5,An1=a3;
        print(An1);*o++=' ';print(An2);*o++='\n';
    }return fwrite(buf,1,o-buf,stdout),0;
}
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