hdu_4547_CD操作(在线LCA)

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题目连接：http://acm.hdu.edu.cn/showproblem.php?pid=4547

题意：中文，不解释

题解：很裸的LCA，注意父目录打开子目录一次就够了，这里我才用倍增在线LCA+map过

#include<cstdio>#include<iostream>#include<map>#include<string>#define F(i,a,b) for(int i=a;i<=b;i++)#pragma comment(linker, "/STACK:1024000000,1024000000")using namespace std;map<string,int>cnt;const int N=1e5+5,DEG=18;int t,n,m,a,b,ed,g[N],nxt[N],v[N],w[N],fa[N][DEG],dep[N],idx;bool fg[N];string s1,s2;inline void adg(int x,int y){v[++ed]=y,nxt[ed]=g[x],g[x]=ed;}int Abs(int a){return a>0?a:-a;}void dfs(int u,int pre){dep[u]=dep[pre]+1,fa[u][0]=pre;F(i,1,DEG-1)fa[u][i]=fa[fa[u][i-1]][i-1];for(int i=g[u];~i;i=nxt[i])if(v[i]!=pre)dfs(v[i],u);}int LCA(int a,int b){if(dep[a]>dep[b])a^=b,b^=a,a^=b;if(dep[a]<dep[b])F(i,0,DEG-1)if((dep[b]-dep[a])&(1<<i))b=fa[b][i];if(a!=b)for(int i=DEG-1;i<0?a=fa[a][0]:0,i>=0;i--)if(fa[a][i]!=fa[b][i])a=fa[a][i],b=fa[b][i];return a;}int main(){scanf("%d",&t);while(t--){scanf("%d%d",&n,&m);cnt.clear();F(i,1,n)g[i]=-1,fg[i]=0;ed=0,idx=1;F(i,1,n-1){cin>>s1>>s2;a=cnt[s1],cnt[s1]=(a==0)?idx++:a,a=(a==0?idx-1:a);b=cnt[s2],cnt[s2]=(b==0)?idx++:b,b=(b==0?idx-1:b);adg(b,a),fg[a]=1;}int root;F(i,1,n)if(!fg[i]){root=i;break;}dep[root]=0;dfs(root,root);F(i,1,m){cin>>s1>>s2;a=cnt[s1],b=cnt[s2];int lca=LCA(a,b),ans=0;if(lca==a&&lca!=b)ans=1;else if(lca==b&&lca!=a)ans=Abs(dep[a]-dep[b]);else if(lca!=a&&lca!=b)ans=Abs(dep[a]-dep[lca])+1;printf("%d\n",ans);}}return 0;}

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