poj_1330 Nearest Common Ancestors（LCA）

Nearest Common Ancestors

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 27746 Accepted: 14240

Description

A rooted tree is a well-known data structure in computer science and engineering. An example is shown below:


In the figure, each node is labeled with an integer from {1, 2,...,16}. Node 8 is the root of the tree. Node x is an ancestor of node y if node x is in the path between the root and node y. For example, node 4 is an ancestor of node 16. Node 10 is also an ancestor of node 16. As a matter of fact, nodes 8, 4, 10, and 16 are the ancestors of node 16. Remember that a node is an ancestor of itself. Nodes 8, 4, 6, and 7 are the ancestors of node 7. A node x is called a common ancestor of two different nodes y and z if node x is an ancestor of node y and an ancestor of node z. Thus, nodes 8 and 4 are the common ancestors of nodes 16 and 7. A node x is called the nearest common ancestor of nodes y and z if x is a common ancestor of y and z and nearest to y and z among their common ancestors. Hence, the nearest common ancestor of nodes 16 and 7 is node 4. Node 4 is nearer to nodes 16 and 7 than node 8 is.

For other examples, the nearest common ancestor of nodes 2 and 3 is node 10, the nearest common ancestor of nodes 6 and 13 is node 8, and the nearest common ancestor of nodes 4 and 12 is node 4. In the last example, if y is an ancestor of z, then the nearest common ancestor of y and z is y.

Write a program that finds the nearest common ancestor of two distinct nodes in a tree.

Input

The input consists of T test cases. The number of test cases (T) is given in the first line of the input file. Each test case starts with a line containing an integer N , the number of nodes in a tree, 2<=N<=10,000. The nodes are labeled with integers 1, 2,..., N. Each of the next N -1 lines contains a pair of integers that represent an edge --the first integer is the parent node of the second integer. Note that a tree with N nodes has exactly N - 1 edges. The last line of each test case contains two distinct integers whose nearest common ancestor is to be computed.

Output

Print exactly one line for each test case. The line should contain the integer that is the nearest common ancestor.

Sample Input

2161 148 510 165 94 68 44 101 136 1510 116 710 216 38 116 1216 752 33 43 11 53 5

Sample Output

43

LCA模板题

在线算法

// dfs+RMQ#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <stack>#include <bitset>#include <queue>#include <set>#include <map>#include <string>#include <algorithm>#define FOP freopen("data.txt","r",stdin)#define FOP2 freopen("data1.txt","w",stdout)#define inf 0x3f3f3f3f#define maxn 10010#define mod 1000000007#define PI acos(-1.0)#define LL long longusing namespace std;struct Edge{    int from, to;    Edge(int u, int v) : from(u), to(v) {}};int n;vector<Edge> edges;vector<int> G[maxn];void addEdge(int from, int to){    edges.push_back(Edge(from, to));    int x = edges.size();    G[from].push_back(x-1);}int father[maxn]; //u的父节点是father[u]int first[maxn]; //记录结点在搜索顺序数组中最先出现的位置（下标）int occur[maxn<<1]; //结点出现的顺序数组int depth[maxn<<1]; //结点在搜索树中的深度，与occur相对应int dp_min[maxn<<1][32]; //dp_min[i][j] 表示从第i个位置开始的2^j个元素中的最小值的下标int cot; void init(){    edges.clear();    for(int i = 1; i <= n; i++) G[i].clear();    cot = 0;    memset(first, 0, sizeof(first));}void dfs(int u, int dep){    occur[++cot] = u;    depth[cot] = dep;    if(!first[u]) first[u] = cot;    for(int i = 0; i < G[u].size(); i++)    {        Edge& e = edges[G[u][i]];        dfs(e.to, dep+1);        occur[++cot] = u;        depth[cot] = dep;    }}void RMQ_init(int n){    for(int i = 1; i <= n; i++) dp_min[i][0] = i;    for(int j = 1; (1<<j) <= n; j++)    {        for(int i = 1; i+(1<<j)-1 <= n; i++)        {            dp_min[i][j] = depth[dp_min[i][j-1]] < depth[dp_min[i+(1<<(j-1))][j-1]] ? dp_min[i][j-1] : dp_min[i+(1<<(j-1))][j-1];        }    }}int RMQ(int L, int R){    L = first[L], R = first[R];    if(L > R) swap(L, R);    int k = 0;    while((1<<(k+1)) <= R-L+1) k++;    int res = depth[dp_min[L][k]] < depth[dp_min[R-(1<<k)+1][k]] ? dp_min[L][k] : dp_min[R-(1<<k)+1][k];    return occur[res]; //取得最小值下标表示的节点}int main(){    int T;    scanf("%d", &T);    while(T--)    {        scanf("%d", &n);        init();        for(int i = 1; i <= n; i++) father[i] = i;        int u, v;        for(int i = 0; i < n-1; i++)        {            scanf("%d%d", &u, &v);            addEdge(u, v);            father[v] = u;        }        for(int i = 1; i <= n; i++)        {            if(father[i] == i) {dfs(i, 0); break; }        }        RMQ_init(cot);        int l, r;        scanf("%d%d", &l, &r);        printf("%d\n", RMQ(l, r));    }    return 0;}

离线算法

// dfs+并查集#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <stack>#include <bitset>#include <queue>#include <set>#include <map>#include <string>#include <algorithm>#define FOP freopen("data.txt","r",stdin)#define FOP2 freopen("data1.txt","w",stdout)#define inf 0x3f3f3f3f#define maxn 10010#define mod 1000000007#define PI acos(-1.0)#define LL long longusing namespace std;struct Edge{    int from, to;    Edge(int u, int v) : from(u), to(v) {}};int n;vector<Edge> edges;vector<int> G[maxn];int father[maxn];int pre[maxn];bool vis[maxn];void addEdge(int from, int to){    edges.push_back(Edge(from, to));    int x = edges.size();    G[from].push_back(x-1);}void init(){    edges.clear();    for(int i = 1; i <= n; i++)    {        G[i].clear();        pre[i] = i;        father[i] = i;        vis[i] = false;    }}int findUF(int x){    if(pre[x] == x) return x;    return pre[x] = findUF(pre[x]);}void joinUF(int u, int v){    u = findUF(u), v = findUF(v);    if(u != v) pre[v] = u; //顺序不能调}int l, r, ans;void LCA(int u){    for(int i = 0; i < G[u].size(); i++)    {        Edge& e = edges[G[u][i]];        LCA(e.to);        joinUF(u, e.to);    }    vis[u] = true;    if(u == l && vis[r] == true)    {        ans = findUF(r);        return ;    }    if(u == r && vis[l] == true)    {        ans = findUF(l);        return ;    }}int main(){    int T;    scanf("%d", &T);    while(T--)    {        scanf("%d", &n);        init();        int u, v;        for(int i = 0; i < n-1; i++)        {            scanf("%d%d", &u, &v);            addEdge(u, v);            father[v] = u;        }        scanf("%d%d", &l, &r);        for(int i = 1; i <= n; i++)        {            if(father[i] == i) { LCA(i); break; }        }        printf("%d\n", ans);    }    return 0;}