POJ1330 Nearest Common Ancestors
Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 17734 Accepted: 9405
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
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
Source
Taejon 2002
tarjan+并查集求LCA模板。将u的子儿子v递归合并到u，若y在x的子树上，则根据并查集的性质，公共祖先即为x（father[y]）。若y和x不再同一子树上。则根据dfs回溯的性质，则搜完y之后一定是回溯到x,y的最近公共祖先才能搜到x的，而此时由并查集的性质，回溯的节点刚好就是father[y]。
#include<map>
#include<set>
#include<stack>
#include<queue>
#include<string>
#include<cstdio>
#include<vector>
#include<cctype>
#include<cassert>
#include<utility>
#include<numeric>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define pr pair
#define MP make_pair
#define VI(x) vector<x>::iterator
#define MI(x,y) map<x,y>::iterator
#define SI(x) set<x>::iterator
#define F first
#define S second
#define clrQ(x) while(!x.empty)x.pop();
#define clrA(x,y) memset(x,y,sizeof(x));
#if defined (_WIN32) || defined (__WIN32) || defined (WIN32) || defined (__WIN32__)
#define LL __int64
#define LLS "%" "I" "6" "4" "d"
#define LLU "%" "I" "6" "4" "u"
#define LL_MAX _I64_MAX
#else
#define LL long long
#define LLS "%" "l" "l" "d"
#define LLU "%" "l" "l" "u"
#define LL_MAX _I64_MAX
#endif
const int inf=~0u>>1;
const LL lnf=~0ull>>1;
#define N 10005
#define M 10005
int n, m;
//UnionSet
int p[N];
void initUset() {
for (int i = 0; i <= n; i++)
p[i] = i;
}
int uFind(int x) {
return x == p[x] ? p[x] : p[x] = uFind(p[x]);
}
void uMerge(int x, int y) { //y合并到x
int fx = uFind(x);
int fy = uFind(y);
if (fx != fy)
p[fy]=fx;
}
//Edge Graph
int head[N], pos;
struct Edge {
int v, nxt;
} e[M];
void initEdge() {
memset(head, -1, sizeof(head));
pos = 0;
}
void add(int u, int v) {
e[pos].v = v;
e[pos].nxt = head[u];
head[u] = pos++;
}
int findRoot(){//有向图选择入度为0的点，无向图不同节点做根，LCA结果将不同。
for (int i = 1; i < n; i++)
if (indeg[i] == 0) return i;
return assert(false),-1;
}
//LCA
int size, root;
int vis[N];
int ans[5];
int indeg[N];
vector<pr<int, int> > que[N];
int dfs(int u) {
int solved=0;
for (int i = head[u]; ~i; i = e[i].nxt) {
int v = e[i].v;
solved+=dfs(v);
if(solved==size)return solved;
uMerge(u, v);
}
vis[u] = 1;
for (int i = 0; i < (int) que[u].size(); i++) {
int v = que[u][i].F;
if (vis[v]) {
ans[que[u][i].S] = uFind(v);
solved++;
}
}
return solved;
}
void tarjan() {
clrA(vis,0);
dfs(findRoot());
}
int main() {
int T;
scanf("%d", &T);
while (T--) {
scanf("%d", &n);
initEdge();
int u, v;
clrA(indeg,0);
for (int i = 1; i < n; i++) {
scanf("%d%d", &u, &v);
add(u, v);
indeg[v]++;
}
initUset();
size = 1;//查询的次数
for(int i=1;i<=n;i++)que[i].clear();
for (int i = 1; i <= size; i++) {
scanf("%d%d", &u, &v);
que[u].push_back(MP(v, i));
que[v].push_back(MP(u, i));
}
tarjan();
for (int i = 1; i <= size; i++) {
printf("%d\n", ans[i]);
}
}
}
#include<map>
#include<set>
#include<cmath>
#include<stack>
#include<queue>
#include<string>
#include<cstdio>
#include<vector>
#include<cctype>
#include<cassert>
#include<utility>
#include<numeric>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define pr pair
#define MP make_pair
#define SI(x) set<x>::iterator
#define VI(x) vector<x>::iterator
#define MI(x,y) map<x,y>::iterator
#define SRI(x) set<x>::reverse_iterator
#define VRI(x) vector<x>::reverse_iterator
#define MRI(x,y) map<x,y>::reverse_iterator
#define F first
#define S second
#define qlr(x) while(!x.empty)x.pop();
#define clr(x,y) memset(x,y,sizeof(x));
#if defined (_WIN32) || defined (__WIN32) || defined (WIN32) || defined (__WIN32__)
#define LL __int64
#define LLS "%" "I" "6" "4" "d"
#define LLU "%" "I" "6" "4" "u"
#define LL_MAX _I64_MAX
#else
#define LL long long
#define LLS "%" "l" "l" "d"
#define LLU "%" "l" "l" "u"
#define LL_MAX _I64_MAX
#endif
const int inf = ~0u >> 1;
const LL lnf = ~0ull >> 1;
#define N 10005
#define M N-1
int n;
int head[N], pos;
struct edge {
int v, nxt;
} e[M];
void add(int u, int v) {
e[pos].v = v;
e[pos].nxt = head[u];
head[u] = pos++;
}
void initEdge() {
clr(head, -1);
pos = 0;
}
int deep[N];
vector<int> dp[N];
int size;
int findRoot() {
for (int i = 1; i <= n; i++) {
if (dp[i].empty()) return i;
}
return -1;
}
void getDeep(int u = findRoot()) {
for (int i = head[u]; ~i; i = e[i].nxt) {
int v = e[i].v;
deep[v] = deep[u] + 1;
getDeep(v);
}
size = max(size, deep[u] + 1);//记录最大深度
}
void initLCA() {
clr(deep, 0);
size = 0;
getDeep();
for (int len = 1,t=2;t<=size; len++,t<<=1) {//跳出条件：(t=2^len)<=size
for (int i = 1; i <= n; i++) {
if ((int)dp[i].size() >= len && (int)dp[dp[i][len-1]].size() >= len) {
dp[i].push_back(dp[dp[i][len-1]][len-1]);
}
}
}
}
int getLCA(int u, int v) {
if (deep[u] < deep[v]) swap(u, v);
int d = deep[u] - deep[v];
for (int i = 0; d; i++, d >>= 1) {
u = d & 1 ? dp[u][i] : u;
}//以差值二进制形式将u往上走差值
if (u == v) return u;
for (int i = min(dp[u].size(),dp[v].size())-1; i >= 0; i--) {
if (dp[u][i] != dp[v][i]) {
u = dp[u][i];
v = dp[v][i];
i=(int)min(dp[u].size(),dp[v].size());//注意每次都要更新i！！
}
}
u = dp[u][0];
return u;
}
int main() {
int T;
cin >> T;
while (T--) {
scanf("%d", &n);
int u, v;
initEdge();
for (int i = 1; i <= n; i++)
dp[i].clear();
for (int i = 1; i < n; i++) {
scanf("%d%d", &u, &v);
add(u, v);
dp[v].push_back(u);
}
initLCA();
scanf("%d%d", &u, &v);
printf("%d\n", getLCA(u, v));
}
}
//============================================================================
// Name : test3.cpp
// Author :
// Version :
// Copyright : Your copyright notice
// Description : Hello World in C++, Ansi-style
//============================================================================
#include<map>
#include<set>
#include<stack>
#include<queue>
#include<string>
#include<cstdio>
#include<vector>
#include<cctype>
#include<cassert>
#include<utility>
#include<numeric>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define pr pair
#define MP make_pair
#define SI(x) set<x>::iterator
#define VI(x) vector<x>::iterator
#define MI(x,y) map<x,y>::iterator
#define SRI(x) set<x>::reverse_iterator
#define VRI(x) vector<x>::reverse_iterator
#define MRI(x,y) map<x,y>::reverse_iterator
#define F first
#define S second
#define clrQ(x) while(!x.empty)x.pop();
#define clrA(x,y) memset(x,y,sizeof(x));
#if defined (_WIN32) || defined (__WIN32) || defined (WIN32) || defined (__WIN32__)
#define LL __int64
#define LLS "%" "I" "6" "4" "d"
#define LLU "%" "I" "6" "4" "u"
#define LL_MAX _I64_MAX
#else
#define LL long long
#define LLS "%" "l" "l" "d"
#define LLU "%" "l" "l" "u"
#define LL_MAX _I64_MAX
#endif
const int inf = ~0u >> 1;
const LL lnf = ~0ull >> 1;
int n, m;
#define N 10005
#define M 10005
int head[N], pos;
struct Edge {
int v, nxt;
} e[M];
void initEdge() {
memset(head, -1, sizeof(head));
pos = 0;
}
void add(int u, int v) {
e[pos].v = v;
e[pos].nxt = head[u];
head[u] = pos++;
}
//RMQ
int dpM[20][N << 1 | 1];
int lg2[N << 1 | 1];
#define getLeft(R,L) (R-(L)+1)
void initRMQ(int n) { //dp[0][i]表示区间i的值。预先处理出来。
lg2[0] = -1;
int limit;
for (int i = 1; i <= n; i++) {
lg2[i] = i & (i - 1) ? lg2[i - 1] : lg2[i - 1] + 1;
}
for (int i = 1; i <= lg2[n]; i++) {
limit = getLeft(n, 1 << i);
for (int j = 1; j <= limit; j++) {
dpM[i][j] = min(dpM[i - 1][j], dpM[i - 1][j + (1 << (i - 1))]);
}
}
}
int getRMQ(int x, int y) {
if (x > y) swap(x, y);
int t = lg2[y - x + 1];
return min(dpM[t][x], dpM[t][getLeft(y, 1 << t)]);
}
//LCA
int depth, cnt;
int inde[N], H[N], E[N];
//dp[0][N<<1|1]深度序列（dfs编号），E[N]每个dfs编号对应的节点，H[N]节点第一次出现在dfs编号序列中的位置
int findRoot() {
for (int i = 1; i <= n; i++)
if (!inde[i]) return i;
return -1;
}
void getEuler(int u = findRoot()) {
int dfn = dpM[0][H[u] = ++cnt] = ++depth;
E[dfn] = u;
for (int i = head[u]; ~i; i = e[i].nxt) {
int v = e[i].v;
getEuler(v);
dpM[0][++cnt] = dfn;
}
}
void initLCA() {
depth = cnt = 0;
getEuler();
initRMQ(cnt);
}
int getLCA(int u, int v) {
if (H[u] > H[v]) swap(u, v);
return E[getRMQ(H[u], H[v])];
}
int main() {
int T;
scanf("%d", &T);
while (T--) {
scanf("%d", &n);m=n-1;
int u, v;
initEdge();
memset(inde, 0, sizeof(inde));
for (int i = 1; i <= m; i++) {
scanf("%d%d", &u, &v);
add(u, v);
inde[v]++;
}
initLCA();
scanf("%d%d", &u, &v);
printf("%d\n", getLCA(u, v));
}
}