Codeforces Round #425 (Div. 2) D. Misha, Grisha and Underground LCA模版

题目链接: Misha, Grisha and Underground

题目大意

给你一颗无根树
三个顶点f, s, t, 计算f到s的最短路径和f到t的最短路径的公共顶点个数
给你三个点a, b, c, 求如何将a, b, c对应到f, s, t, 使得上面描述的公共顶点个数最多, 输出最多的公共顶点个数

思路

将顶点1定义为树根, 求出所有顶点到树根的最短路长度depth[i]
那么两个顶点u, v的最短路长度为dis(u, v) = depth[u] + depth[v] - depth[lca(u, v)]*2, lca(u, v)为uv的最近公共祖先
那么f到s和f到t的最短路径的公共边数量就等于(dis(f, s)+dis(f, t)-dis(s, t))/2公共顶点数量还要再+1

代码

第一份代码用的基于二分的LCA, O(nlogn)预处理, O(logn)查询
第二份代码用的是基于RMQ(ST表)的LCA, (nlogn)预处理, O(1)查询

基于二分的LCA

GNU C++14 Accepted 327 ms 18100 KB

#include <bits/stdc++.h>using namespace std;const int maxv = 1e5 + 100, maxlog = 17;vector<int> G[maxv];int root, par[maxlog][maxv];int depth[maxv];//预处理O(nlogn), 查询log(n)void dfs(int v, int p, int d){    par[0][v] = p;    depth[v] = d;    for(auto &ite : G[v]) if(ite!=p) dfs(ite, v, d+1);}void init(int V){    dfs(root, -1, 0);    for(int k=0; k+1<maxlog; ++k)    {        for(int v=1; v<=V; ++v)        {            if(par[k][v] < 0) par[k+1][v] = -1;            else par[k+1][v] = par[k][par[k][v]];        }    }}int lca(int u, int v){    if(depth[u] > depth[v]) swap(u, v);    for(int k=0; k<maxlog; ++k)    {        if(((depth[v]-depth[u]) >> k) & 1) v = par[k][v];    }    if(u == v) return u;    for(int k=maxlog-1; k>=0; --k)    {        if(par[k][u] != par[k][v])        {            u = par[k][u];            v = par[k][v];        }    }    return par[0][u];}void add(int u, int v){    G[u].push_back(v);    G[v].push_back(u);}const int maxn = 1e5 + 10;int n, q, u, v;int dis(int u, int v){    return depth[u] + depth[v] - depth[lca(u, v)]*2;}int cal(int f, int s, int t){    return (dis(f, s)+dis(f, t)-dis(s, t))/2;}int main(){    scanf("%d%d", &n, &q);    for(int i=2; i<=n; ++i)    {        scanf("%d", &u);        add(u, i);    }    root = 1;    init(n);    int a, b, c;    while(q--)    {        scanf("%d%d%d", &a, &b, &c);        printf("%d\n", max(cal(a, b, c), max(cal(b, a, c), cal(c, a, b))) + 1);    }    return 0;}

基于RMQ的LCA

GNU C++14 Accepted 264 ms 36500 KB

#include <bits/stdc++.h>using namespace std;const int maxv = 1e5+100;vector<int> G[maxv];int root, vs[maxv*2], depth[maxv*2], id[maxv], mi[maxv*2][30];void init_rmq(int n){    int a, b;    for(int i=1; i<=n; ++i) mi[i][0] = i;    for(int j=1; (1<<j)<=n; ++j)    {        for(int i=1; i+(1<<j)-1<=n; ++i)        {            a = mi[i][j-1], b = mi[i+(1<<(j-1))][j-1];            mi[i][j] = depth[a]<depth[b] ? a : b;        }    }}int rmq(int l, int r){    int k = 0;    while((1<<(k+1)) <= r-l+1) ++k;    int a = mi[l][k], b = mi[r-(1<<k)+1][k];    return depth[a] < depth[b] ? a : b;}void dfs(int v, int p, int d, int &k){    id[v] = k;    vs[k] = v;    depth[k++] = d;    for(auto &ite : G[v])    {        if(ite!=p)         {            dfs(ite, v, d+1, k);            vs[k] = v;            depth[k++] = d;        }    }}void init(int V){    int k = 1;    dfs(root, -1, 0, k);    init_rmq(V*2-1);}int lca(int u, int v){    if(id[u] > id[v]) swap(u, v);    return vs[rmq(id[u], id[v])];}void add(int u, int v){    G[u].push_back(v);    G[v].push_back(u);}const int maxn = 1e5 + 10;int n, q, u, v;int dis(int u, int v){    return depth[id[u]] + depth[id[v]] - depth[id[lca(u, v)]]*2;}int cal(int f, int s, int t){    return (dis(f, s)+dis(f, t)-dis(s, t))/2;}int main(){    scanf("%d%d", &n, &q);    for(int i=2; i<=n; ++i)    {        scanf("%d", &u);        add(u, i);    }    root = 1;    init(n);    int a, b, c;    while(q--)    {        scanf("%d%d%d", &a, &b, &c);        printf("%d\n", max(cal(a, b, c), max(cal(b, a, c), cal(c, a, b))) + 1);    }    return 0;}