A. Nearest Common Ancestors

A. Nearest Common Ancestors

Time Limit: 1000ms

Case Time Limit: 1000ms

Memory Limit: 10000KB

 

64-bit integer IO format: %lld      Java class name: Main

 

 

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

2
16
1 14
8 5
10 16
5 9
4 6
8 4
4 10
1 13
6 15
10 11
6 7
10 2
16 3
8 1
16 12
16 7
5
2 3
3 4
3 1
1 5
3 5

 

Sample Output

4
3

解题：本来打算用dfs+RMQ写的，结果写得一塌糊涂，完全不对。后来发现只有一个询问，即为了这一次询问可以破坏原有结构。直接暴力好了。

#include <iostream>
#include <cstdio>
using namespace std;
bool vis[10005];
int uf[10005];
int main() {
    int ks,n,i,x,y;
    scanf("%d",&ks);
    while(ks--) {
        scanf("%d",&n);
        for(i = 0; i <= n; i++) {
            vis[i] = false;
            uf[i] = i;
        }
        for(i = 1; i < n; i++) {
            scanf("%d %d",&x,&y);
            uf[y] = x;
        }
        scanf("%d %d",&x,&y);
        vis[x] = true;
        x = uf[x];
        while(x != uf[x]) {
            vis[x] = true;
            x = uf[x];
        }
        while(y != uf[y]) {
            if(vis[y]) break;
            y = uf[y];
        }
        printf("%d
",y);
    }
    return 0;
}

倍增法求LCA

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <climits>
#include <vector>
#include <queue>
#include <cstdlib>
#include <string>
#include <set>
#include <stack>
#define LL long long
#define pii pair<int,int>
#define INF 0x3f3f3f3f
using namespace std;
const int maxn = 10010;
int n,deep[maxn],fa[maxn][32];
vector<int>g[maxn];
void dfs(int u,int f){
    for(int i = 0; i < g[u].size(); i++){
        if(g[u][i] == f) continue;
        deep[g[u][i]] = deep[u]+1;
        dfs(g[u][i],u);
    }
}
void init(){
    for(int j = 1; j < 32; j++){
        for(int i = 1; i <= n; i++)
            fa[i][j] = fa[fa[i][j-1]][j-1];
    }
}
int LCA(int u,int v){
    if(deep[u] < deep[v]) swap(u,v);
    int i,d = deep[u]-deep[v];
    for(i = 0; i < 32; i++) if((1<<i)&d) u = fa[u][i];
    if(u == v) return u;
    for(i = 31; i >= 0; i--){
        if(fa[u][i] != fa[v][i]){
            u = fa[u][i];
            v = fa[v][i];
        }
    }
    return fa[u][0];
}
int main() {
    int t,i,u,v,root;
    scanf("%d",&t);
    while(t--){
        scanf("%d",&n);
        for(i = 0; i <= n; i++){
            g[i].clear();
            deep[i] = 0;
        }
        memset(fa,0,sizeof(fa));
        for(i = 1; i < n; i++){
            scanf("%d %d",&u,&v);
            g[u].push_back(v);
            g[v].push_back(u);
            fa[v][0] = u;
            if(!fa[u][0]) root = u;
        }
        deep[root] = 1;
        dfs(root,-1);
        init();
        scanf("%d %d",&u,&v);
        printf("%d
",LCA(u,v));
    }
    return 0;
}

tarjan求LCA

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <climits>
#include <vector>
#include <queue>
#include <cstdlib>
#include <string>
#include <set>
#define LL long long
#define INF 0x3f3f3f3f
using namespace std;
const int maxn = 10010;
vector<int>g[maxn];
int n,x,y,uf[maxn],ans;
bool vis[maxn];
int Find(int rt) {
    if(rt != uf[rt])
        uf[rt] = Find(uf[rt]);
    return uf[rt];
}
bool dfs(int u) {
    uf[u] = u;
    for(int i = 0; i < g[u].size(); i++) {
        if(!vis[g[u][i]]) {
            if(dfs(g[u][i])) return true;
            uf[g[u][i]] = u;
        }
    }
    if(u == x && vis[y]) {
        ans = Find(y);
        return true;
    } else if(u == y && vis[x]) {
        ans = Find(x);
        return true;
    }
    vis[u] = true;
    return false;
}
int main() {
    int t,i,u,v,root;
    scanf("%d",&t);
    while(t--) {
        scanf("%d",&n);
        for(i = 0; i <= n; i++) {
            g[i].clear();
            vis[i] = false;
        }
        for(i = 1; i < n; i++) {
            scanf("%d %d",&u,&v);
            g[u].push_back(v);
            vis[v] = true;
        }
        for(i = 1; i <= n; i++)
            if(!vis[i]) {
                root = i;
                break;
            }
        scanf("%d%d",&x,&y);
        memset(vis,false,sizeof(vis));
        dfs(root);
        cout<<ans<<endl;
    }
    return 0;
}