一、前人种树
博客:浅谈倍增法求LCA
二、沙场练兵
题目:POJ 1330 Nearest Common Ancestors
代码:
const int MAXN = 10010;
const int DEG = 20; struct Edge
{
int to,next;
}edge[MAXN*2];
int head[MAXN],tot;
void addedge(int u,int v)
{
edge[tot].to = v;
edge[tot].next = head[u];
head[u] = tot++;
}
void init()
{
tot = 0;
memset(head,-1,sizeof(head));
}
int fa[MAXN][DEG];//fa[i][j]表示结点i的第2^j个祖先
int deg[MAXN];//深度数组 void BFS(int root)
{
queue<int>que;
deg[root] = 0;
fa[root][0] = root;
que.push(root);
while(!que.empty())
{
int tmp = que.front();
que.pop();
for(int i = 1;i < DEG;i++)
fa[tmp][i] = fa[fa[tmp][i-1]][i-1];
for(int i = head[tmp]; i != -1;i = edge[i].next)
{
int v = edge[i].to;
if(v == fa[tmp][0])continue;
deg[v] = deg[tmp] + 1;
fa[v][0] = tmp;
que.push(v);
} }
}
int LCA(int u,int v)
{
if(deg[u] > deg[v])swap(u,v);
int hu = deg[u], hv = deg[v];
int tu = u, tv = v;
for(int det = hv-hu, i = 0; det ;det>>=1, i++)
if(det&1)
tv = fa[tv][i];
if(tu == tv)return tu;
for(int i = DEG-1; i >= 0; i--)
{
if(fa[tu][i] == fa[tv][i])
continue;
tu = fa[tu][i];
tv = fa[tv][i];
}
return fa[tu][0];
}
bool flag[MAXN];
int main()
{
int T;
int n;
int u,v;
scanf("%d",&T);
while(T--)
{
scanf("%d",&n);
init();
memset(flag,false,sizeof(flag));
for(int i = 1;i < n;i++)
{
scanf("%d%d",&u,&v);
addedge(u,v);
addedge(v,u);
flag[v] = true;
}
int root;
for(int i = 1;i <= n;i++)
if(!flag[i])
{
root = i;
break;
}
BFS(root);
scanf("%d%d",&u,&v);
printf("%d\n",LCA(u,v));
}
return 0;
}