LCA问题

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LCA问题

求树上的最近公共祖先

1 倍增法

倍增法，算法正如它的名字一样，比较好理解，预处理出parent[k][u] ，表示从u节点向上2^k步到大的节点编号，递推的时候

$$parent[k+1][u] = parent[k][parent[k][u]]$$
如果需要求距离
$$dis[k+1][u] =dis[k][u]+ dis[k][parent[k][u]$$

POJ 1330

// POJ 1330
const int maxn = 1e4+100;
const int maxlogv = 14;
vector<int> G[maxn];
int root;

int parent[maxlogv][maxn];
int depth[maxn];

void dfs(int v,int p,int d){
    parent[0][v] = p;
    depth[v] = d;
    for(int i = 0;i < G[v].size(); ++i){
        if(G[v][i] != p){
            dfs(G[v][i],v,d+1);
        }
    }
}
void init(int V){

    dfs(root,-1,0);
    for(int k = 0;k+1 < maxlogv; ++k){
        for(int v = 0; v < V; ++v){
            if(parent[k][v] < 0) parent[k+1][v] = -1;
            else parent[k+1][v] =  parent[k][parent[k][v]];

        }
    }
}

int lca(int u,int v){
    if(depth[u] > depth[v]) swap(u,v);
    for(int k = 0;k < maxlogv; ++k){
        if(((depth[v] - depth[u]) >> k)& 1){
            v = parent[k][v];
        }

    }
    // cout<<depth[u]<<" "<<depth[v]<<endl;
    // cout<<v+1<<" "<<u+1<<endl;
    if(u == v) return u;
    for(int k = maxlogv-1; k >= 0; --k){
        if(parent[k][u] != parent[k][v]){
            u = parent[k][u];
            v = parent[k][v];
        }
    }
    return parent[0][u];
}
bool OUT[maxn];
int main(void)
{

    int T;
    scanf("%d",&T);
    while(T--){
        int n;
        rep(i,0,n) G[i].clear();
        me(OUT);
        scanf("%d",&n);
        // cout<<n<<endl;
        rep(i,1,n) {
            int u,v;
            scanf("%d %d",&u,&v);
            u--,v--;
            G[u].push_back(v);
            OUT[v] = 1;
            // G[v].push_back(u);
        }
        rep(i,0,n) if(!OUT[i]){
            root = i;
            break;
        }
        // cout<<root<<endl;        // root = 0;
        init(n);

        int u,v;

        scanf("%d %d",&u,&v);
        u--,v--;
        printf("%d\n",lca(u,v)+1);
    }

   return 0;
}

HDU2586

HDU2586 求树上的最短路，LCA+倍增
typedef pair<int,int> P;

const int maxn = 4e4+100;
const int maxlogv = 16;
vector<P> G[maxn];
int root;

int parent[maxlogv][maxn];
int depth[maxn];
int dis[maxlogv][maxn];
void dfs(int v,int p,int d,int D){
    parent[0][v] = p;
    dis[0][v] = D;
    depth[v] = d;
    for(int i = 0;i < G[v].size(); ++i){
        P &to = G[v][i];
        if(to.FI != p){
            dfs(to.FI,v,d+1,to.SE);
        }
    }
}
void init(int V){
    memset(dis,0,sizeof(dis));
    dfs(root,-1,0,0);
    for(int k = 0;k+1 < maxlogv; ++k){
        for(int v = 0; v < V; ++v){
            if(parent[k][v] < 0) {
                parent[k+1][v] = -1;
                // dis[k+1][v] = dis[k][v];
            }
            else {
                dis[k+1][v] = dis[k][v] + dis[k][parent[k][v]];
                parent[k+1][v] =  parent[k][parent[k][v]];
                }
        }
    }
}

int lca(int u,int v){
    int ans = 0;
    if(depth[u] > depth[v]) swap(u,v);
        // cout<<depth[u]<<" "<<depth[v]<<endl;
    for(int k = 0;k < maxlogv; ++k){
        if(((depth[v] - depth[u]) >> k)& 1){
            ans += dis[k][v];
            v = parent[k][v];
        }
    }

    // cout<<v+1<<" "<<u+1<<endl;
    if(u == v) return ans;
    // cout<<u<<" "<<v<<endl;
    for(int k = maxlogv-1; k >= 0; --k){
        if(parent[k][u] != parent[k][v]){
            ans += dis[k][u];
            ans += dis[k][v];
            u = parent[k][u];
            v = parent[k][v];

        }
    }
    if(u != v) ans += dis[0][u]+dis[0][v];
    return ans;
}
int main(void)
{

    int T;
    scanf("%d",&T);
    // cout<<T<<endl;
    while(T--){
        int n,m;
        scanf("%d %d",&n,&m);
        rep(i,0,n) G[i].clear();

        int u,v,w;
        rep(i,1,n) {
            scanf("%d %d %d",&u,&v,&w);
            u--,v--;
            G[u].push_back(P(v,w));
            G[v].push_back(P(u,w));
        }
        root = 0;
        init(n);
        // cout<<dis[0][1]<<endl;
        // cout<<dis[0][2]<<endl;
        while(m--){
            scanf("%d %d",&u,&v);
            u--,v--;
            // cout<<u<<" "<<v<<endl;
            // cout<<depth[u]<<" "<<depth[v]<<endl;
            printf("%d\n",lca(u,v));
        }
    }

   return 0;
}