hihoCoder 1167高等理论计算机科学(LCA，没有完全理解，先放着)

clj在某场hihoCoder比赛中的一道题，表示clj的数学题实在6，这道图论貌似还算可以。。。

题目链接：http://hihocoder.com/problemset/problem/1167

由于是中文题目，题意不再赘述。

对于任意两条小精灵的活动路径a和b，二者相交的判断条件为b的两个端点的LCA在a的路径上；那么我们可以首先将每个活动路径端点的LCA离线预处理出来，对每个节点LCA值+1。

然后以某个节点(我选择的是节点1)为根进行深搜，算出一条从节点1到节点x的LCA值和，那么任意路径a(假设其两端点分别是A和B)上的节点个数就是sum[A] + sum[B] - 2 * sum[LCA(A,B)]。

最后，对于某些点，如果它是不止一条路径的LCA，那么我们只需要对最终答案乘以C(LCAnum, 2)的组合数就好。

【PS：clj给出的题解中，采用了点分治+LCA的方式，虽然看懂了题意，但是表示对递归分治之后的路径，如何求出其上的LCAnum，并没有多好的想法，还望巨巨能指点一下，Thx~】

AC代码：

 

#include <cstdio>
#include <iostream>
#include <cstring>
using  namespace  std;
typedef  long  long  LL;
#define MAXN 100010
struct  Edge {
    int  to, next;
} edge[MAXN << 1];
struct  Node {
    int  to, next, num;
} Query[MAXN << 1];
struct  node {
    int  u, v, lca;
} input[MAXN];
int  totEdge, totQuery, n, m;
int  headEdge[MAXN], headQuery[MAXN];
int  ancestor[MAXN], father[MAXN], LCAnum[MAXN], sum[MAXN];
bool  vis[MAXN];
void  addEdge( int  from,  int  to) {
    edge[totEdge].to = to;
    edge[totEdge].next = headEdge[from];
    headEdge[from] = totEdge++;
}
void  addQuery( int  from,  int  to,  int  x) {
    Query[totQuery].to = to;
    Query[totQuery].num = x;
    Query[totQuery].next = headQuery[from];
    headQuery[from] = totQuery++;
}
void  init() {
    memset (headEdge, -1,  sizeof (headEdge));
    memset (headQuery, -1,  sizeof (headQuery));
    memset (father, -1,  sizeof (father));
    memset (vis,  false ,  sizeof (vis));
    memset (sum, 0,  sizeof (sum));
    memset (LCAnum, 0,  sizeof (LCAnum));
    totEdge = totQuery = 0;
}
int  find_set( int  x) {
    if (x == father[x])  return  x;
    else  return  father[x] = find_set(father[x]);
}
void  union_set( int  x,  int  y) {
    x = find_set(x); y = find_set(y);
    if (x != y) father[y] = x;
}
void  Tarjan( int  u) {
    father[u] = u;
    for ( int  i = headEdge[u]; i != -1; i = edge[i].next) {
        int  v = edge[i].to;
        if (father[v] != -1)  continue ;
        Tarjan(v);
        union_set(u, v);
    }
    for ( int  i = headQuery[u]; i != -1; i = Query[i].next) {
        int  v = Query[i].to;
        if (father[v] == -1)  continue ;
        input[Query[i].num].lca = find_set(v);
    }
}
void  DFS( int  u,  int  pre) {
    vis[u] = 1;
    sum[u] = sum[pre] + LCAnum[u];
    for ( int  i = headEdge[u]; i != -1; i = edge[i].next) {
        int  v = edge[i].to;
        if (vis[v])  continue ;
        DFS(v, u);
    }
}
int  main() {
    init();
    scanf ( "%d%d" , &n, &m);
    for ( int  i = 0; i < n - 1; i++) {
        int  a, b;
        scanf ( "%d%d" , &a, &b);
        addEdge(a, b); addEdge(b, a);
    }
    for ( int  i = 0; i < m; i++) {
        int  a, b;
        scanf ( "%d%d" , &a, &b);
        input[i].u = a, input[i].v = b;
        addQuery(a, b, i); addQuery(b, a, i);
    }
    Tarjan(1);
    for ( int  i = 0; i < m; i++)
        LCAnum[input[i].lca]++;
    DFS(1, 0);
    LL ans = 0;
    for ( int  i = 0; i < m; i++) {
        ans += (sum[input[i].u] + sum[input[i].v] - 2 * sum[input[i].lca]);
    }
    for ( int  i = 1; i <= n; i++) {
        ans += (LL)LCAnum[i] * (LCAnum[i] - 1) / 2;
    }
    printf ( "%lld\n" , ans);
    return  0;
}