Codeforces Gym-101161E【LCA+主席树】

思路：
在这棵树上，对于每个节点u存一下从根节点（默认为1） 到 u 路径上所有线段出现的次数，那么用主席树维护所有历史信息，然后计算任意两个特定节点u, v的所有线段次数：num[线段树（u）] + num[线段树（v）] - 2 * num[线段树（LCA(u, v)）]，然后对于每次询问询问主席树就好了，就是求区间中位数了嘛

代码来源：http://www.cnblogs.com/fightfordream/p/7159024.html

#include<cstdio>#include<vector>#include<string.h>#include<algorithm>using namespace std;#define mem(a,b) memset(a, b, sizeof(a))#define Lson num<<1,Left,Mid#define Rson num<<1|1, Mid+1, Rightconst int Maxn = 1e5 + 7;struct Edge{int to,nex,w;}edge[Maxn<<1];struct Node{int Left,Right,sum;}T[Maxn*40];int n,m,mx,cas,dis[Maxn],root[Maxn],cnt,dp[Maxn][25],dep[Maxn],fa[Maxn],head[Maxn],tol;void Add(int u, int v, int w){    edge[tol] = (Edge){v,head[u],w}, head[u] = tol++;    edge[tol] = (Edge){u,head[v],w}, head[v] = tol++;}void update(int pre, int &rt, int Left, int Right, int pos){    T[++cnt] = T[pre], T[cnt].sum++, rt = cnt;    if(Left == Right) return;    int Mid = (Left + Right) >> 1;    if(Mid >= pos) update(T[pre].Left, T[rt].Left, Left, Mid, pos);    else update(T[pre].Right, T[rt].Right, Mid + 1, Right, pos);}int query(int Left, int Right, int f, int x, int y, int k){    if(x == y) return x;    int Mid = (x + y) >> 1;    int temp = T[T[Left].Left].sum + T[T[Right].Left].sum - T[T[f].Left].sum*2;    if(k <= temp) return query(T[Left].Left, T[Right].Left, T[f].Left, x, Mid, k);    return query(T[Left].Right, T[Right].Right, T[f].Right, Mid + 1, y, k - temp);}void dfs(int u, int f){    dp[u][0] = fa[u];    for(int i = 1; i <= 20; i++) dp[u][i] = dp[dp[u][i-1]][i-1];    for(int i = head[u]; ~i; i = edge[i].nex){        int v = edge[i].to, w = edge[i].w;        if(v == f) continue;        fa[v] = u; dep[v] = dep[u] + 1; dis[u] = dis[v] + w;        update(root[u], root[v], 1, mx, w);        dfs(v, u);    }}int LCA(int x, int y) {    if(dep[x] < dep[y]) swap(x, y);    for(int i = 20; i >= 0; i--)        if(dep[dp[x][i]] >= dep[y]) x = dp[x][i];    if(x == y) return x;    for(int i = 20; i >= 0; i--)        if(dp[x][i] != dp[y][i]) x = dp[x][i], y = dp[y][i];    return dp[x][0];}int main(){    int u,v,w,q,a,b,f,p;    double ans;    scanf("%d", &cas);    while(cas--){        scanf("%d", &n);        tol = 0;        mx = 0;        cnt = 0;        mem(head,-1);        mem(dp,0);        mem(dis,0);        mem(dep,0);        for(int i=1;i<n;i++){            scanf("%d%d%d",&u,&v,&w);            Add(u,v,w);            mx = mx > w ? mx : w;        }        fa[1] = 1;        dfs(1, 0);        scanf("%d",&q);        while(q--){            scanf("%d%d",&a,&b);            f = LCA(a, b);            p = dep[a] + dep[b] - (dep[f]<<1);            ans = 0;            if(p % 2) ans = (double)query(root[a], root[b], root[f], 1, mx, p / 2 + 1);            else ans = ((double)query(root[a], root[b], root[f], 1, mx, p / 2) + (double)query(root[a], root[b], root[f], 1, mx, p / 2 + 1)) / 2.0;            printf("%.1f\n", ans);        }    }    return 0;}