传送门:POJ 1986: Distance Queries
题意:
LCA距离:快速查询树中任意两个节点间的最短距离。
思路:
树中,每两个结点的路径是唯一的,比POJ: 2763简单,稍微改改就过了,思路一致,解题套路。
代码如下:
import java.io.BufferedReader;
import java.io.File;
import java.io.FileInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
import java.util.Map;
import java.util.StringTokenizer;
public class Main{
String INPUT = "./data/judge/201709/P1986.txt";
public static void main(String[] args) throws IOException {
new Main().run();
}
int N, M;
int root; // 选定树的任意一个根
List<Edge>[] g; // 树的邻接表
int[] id; // 记录每个结点最先访问的时间戳
int[] vs; // dfs访问顺序
int[] dp; // 访问过程中对应结点的深度
void init(int n) {
g = new ArrayList[n];
for (int i = 0; i < n; ++i) g[i] = new ArrayList<Edge>();
id = new int[n * 2];
vs = new int[n * 2];
dp = new int[n * 2];
}
void add(int from, int to, int cost) {
g[from].add(new Edge(to, cost));
g[to].add(new Edge(from, cost));
}
class Edge{
int to;
int cost;
Edge(int to, int cost){
this.to = to;
this.cost = cost;
}
}
class RMQ {
int n_;
int[] dat;
int[] data;
RMQ (int N) {
n_ = 1;
while (n_ < N) n_ *= 2;
dat = new int[2 * n_];
for (int i = 0; i < 2 * n_ - 1; ++i) dat[k] = -1;
}
RMQ (int[] data, int N) {
this(N);
this.data = data;
for (int i = 0; i < N; ++i) {
update(i, i);
}
}
void update(int k, int i) {
k += (n_ - 1);
dat[k] = i;
while (k > 0) {
k = (k - 1) / 2;
dat[k] = min(dat[2 * k + 1], dat[2 * k + 2]);
}
}
int min(int i, int j) { // -1表示无穷大
if (i == -1 && j != -1) return j;
if (j == -1 && i != -1) return i;
if (j == -1 && i == -1) return -1;
return data[i] < data[j] ? i : j;
}
int query(int k, int i, int j, int l, int r) {
if (j <= l || i >= r) return -1;
else if (i <= l && j >= r) {
return dat[k];
}
else {
int lch = 2 * k + 1;
int rch = 2 * k + 2;
int mid = (l + r) / 2;
int lf = query(lch, i, j, l, mid);
int rt = query(rch, i, j, mid, r);
return min(lf, rt);
}
}
}
class BIT{
int n;
int[] BIT;
BIT(int n){
this.n = n;
BIT = new int[n + 16];
}
void add(int i, int val) {
while (i <= n) {
BIT[i] += val;
i += i & -i;
}
}
long sum(int i) {
long s = 0;
while (i > 0) {
s += BIT[i];
i -= i & -i;
}
return s;
}
}
BIT bit;
RMQ rmq;
void solve() {
bit = new BIT(2 * N);
k = 0;
dfs(root, -1, 0);
rmq = new RMQ(dp, 2 * N);
}
int k = 0;
void dfs(int v, int p, int d) {
id[v] = k;
vs[k] = v;
dp[k++] = d;
for (Edge e : g[v]) {
if (e.to != p) {
bit.add(k, e.cost);
dfs(e.to, v, d + 1);
vs[k] = v;
bit.add(k, -e.cost);
dp[k++] = d;
}
}
}
int lca(int u, int v) {
return vs[rmq.query(0, Math.min(id[u], id[v]), Math.max(id[u], id[v]) + 1, 0, rmq.n_)];
}
void read() {
N = ni();
M = ni();
root = N / 2;
init(N);
for (int i = 0; i < M; ++i) {
int from = ni();
int to = ni();
int cost = ni();
char r = nc();
from --;
to --;
add(from, to, cost);
}
solve();
int q = ni();
for (int i = 0; i < q; ++i) {
int u = ni();
int v = ni();
u --;
v --;
int p = lca(u, v);
long ans = bit.sum(id[u]) + bit.sum(id[v]) - 2 * bit.sum(id[p]);
out.println(ans);
}
}
FastScanner in;
PrintWriter out;
void run() throws IOException {
boolean oj;
try {
oj = ! System.getProperty("user.dir").equals("F:\\java_workspace\\leetcode");
} catch (Exception e) {
oj = System.getProperty("ONLINE_JUDGE") != null;
}
InputStream is = oj ? System.in : new FileInputStream(new File(INPUT));
in = new FastScanner(is);
out = new PrintWriter(System.out);
long s = System.currentTimeMillis();
read();
out.flush();
if (!oj){
System.out.println("[" + (System.currentTimeMillis() - s) + "ms]");
}
}
public boolean more(){
return in.hasNext();
}
public int ni(){
return in.nextInt();
}
public long nl(){
return in.nextLong();
}
public double nd(){
return in.nextDouble();
}
public String ns(){
return in.nextString();
}
public char nc(){
return in.nextChar();
}
class FastScanner {
BufferedReader br;
StringTokenizer st;
boolean hasNext;
public FastScanner(InputStream is) throws IOException {
br = new BufferedReader(new InputStreamReader(is));
hasNext = true;
}
public String nextToken() {
while (st == null || !st.hasMoreTokens()) {
try {
st = new StringTokenizer(br.readLine());
} catch (Exception e) {
hasNext = false;
return "##";
}
}
return st.nextToken();
}
String next = null;
public boolean hasNext(){
next = nextToken();
return hasNext;
}
public int nextInt() {
if (next == null){
hasNext();
}
String more = next;
next = null;
return Integer.parseInt(more);
}
public long nextLong() {
if (next == null){
hasNext();
}
String more = next;
next = null;
return Long.parseLong(more);
}
public double nextDouble() {
if (next == null){
hasNext();
}
String more = next;
next = null;
return Double.parseDouble(more);
}
public String nextString(){
if (next == null){
hasNext();
}
String more = next;
next = null;
return more;
}
public char nextChar(){
if (next == null){
hasNext();
}
String more = next;
next = null;
return more.charAt(0);
}
}
}