hdu 4547(LCA+Tarjan)

解题思路：很明显的LCA问题，用Tarjan离线算法即可。这里输入的可能是字符串，所以直接用map保存。此外，根据题意，这里需要稍稍有点变化，因为cd：a\b\c...这里是一步即可完成，所以在查询a和b时，还要判断与公共祖先的关系。还要注意，这道题没有告诉根节点，所以可以根据入度为0来判断根节点。

#include<iostream>
#include<string>
#include<cstdio>
#include<cstring>
#include<map>
using namespace std;

const int maxn = 100005;
const int WHITE = 0;
const int GRAY = 1;
const int BLACK = 2;
struct Edge
{
int to,next;
}edge[maxn];
struct Ask
{
int to,next,id,lca;
}ask[maxn<<1];
struct Query
{
int a,b;
}q[maxn];
int n,m,cnt,cnt1,cnt2,pre[maxn],head[maxn];
int fa[maxn],color[maxn],dis[maxn],in[maxn],lca[maxn];
bool vis[maxn];
map<string,int> mp;

void init()
{
mp.clear();
memset(pre,-1,sizeof(pre));
memset(head,-1,sizeof(head));
memset(in,0,sizeof(in));
memset(color,0,sizeof(color));
cnt = cnt1 = cnt2 = 0;
}

void addedge(int u,int v)
{
edge[cnt1].to = v;
edge[cnt1].next = pre[u];
pre[u] = cnt1++;
}

void addask(int u,int v,int id)
{
ask[cnt2].to = v;
ask[cnt2].id = id;
ask[cnt2].next = head[u];
head[u] = cnt2++;
}

int find(int x)
{
if(fa[x] == x) return x;
return fa[x] = find(fa[x]);
}

void Tarjan(int u,int dep)
{
fa[u] = u;
dis[u] = dep;
color[u] = GRAY;
for(int i = pre[u]; i != -1; i = edge[i].next)
{
int v = edge[i].to;
Tarjan(v,dep+1);
fa[v] = u;
}
color[u] = BLACK;
for(int i = head[u]; i != -1; i = ask[i].next)
{
int v = ask[i].to;
if(color[u] == BLACK)
{
int ancestor = find(v);
lca[ask[i].id] = ancestor;
}
}
}

int main()
{
int t;
string A,B;
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&n,&m);
init();
for(int i = 1; i < n; i++)
{
cin >> A >> B;
if(mp.find(A) == mp.end())
mp[A] = cnt++;
if(mp.find(B) == mp.end())
mp[B] = cnt++;
addedge(mp[B],mp[A]);
in[mp[A]]++;
}
for(int i = 1; i <= m; i++)
{
cin >> A >> B;
addask(mp[A],mp[B],i);
addask(mp[B],mp[A],i);
q[i].a = mp[A], q[i].b = mp[B];
}
int root;
for(int i = 0; i < n; i++)
if(in[i] == 0)
{
root = i;
break;
}
Tarjan(root,0);
for(int i = 1; i <= m; i++)
{
int ancestor = lca[i];
int a = q[i].a;
int b = q[i].b;
if(a == b)
printf("0\n");
else if(a == ancestor)
printf("1\n");
else if(b == ancestor)
printf("%d\n",dis[a] - dis[b]);
else printf("%d\n",dis[a] - dis[ancestor] + 1);
}
}
return 0;
}