题意:
求结点大于等于2的强连通分量个数。
直接上代码,没什么好说的。
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.StringTokenizer;
public class Main{
String INPUT = "./data/judge/201709/P3180.txt";
public static void main(String[] args) throws IOException {
new Main().run();
}
static final int MAX_N = 10000 + 16;
List<Edge>[] g, rg;
List<Integer> po;
boolean[] used;
int[] cmp;
int V;
class Edge{
int from;
int to;
Edge(int from, int to){
this.from = from;
this.to = to;
}
}
void init(int V) {
this.V = V;
g = new ArrayList[MAX_N];
rg = new ArrayList[MAX_N];
po = new ArrayList<Integer>();
used = new boolean[MAX_N];
cmp = new int[MAX_N];
Arrays.fill(used, false);
for (int i = 0; i < V; ++i) g[i] = new ArrayList<Edge>();
for (int i = 0; i < V; ++i) rg[i] = new ArrayList<Edge>();
}
void add(int from, int to) {
g[from].add(new Edge(from, to));
rg[to].add(new Edge(to, from));
}
void dfs(int v) {
used[v] = true;
for (Edge e : g[v]) {
if (!used[e.to]) dfs(e.to);
}
po.add(v);
}
void rdfs(int v, int k) {
used[v] = true;
cmp[v] = k;
for (Edge e : rg[v]) {
if (!used[e.to]) rdfs(e.to, k);
}
}
int kosarajuSCC() {
for (int i = 0; i < V; ++i) {
if (!used[i]) dfs(i);
}
Arrays.fill(used, false);
int k = 0;
for (int i = po.size() - 1; i >= 0; --i) {
if (!used[po.get(i)]) rdfs(po.get(i), k++);
}
return k;
}
void read() {
int N = ni();
int M = ni();
init(N);
for (int i = 0; i < M; ++i) {
int from = ni();
int to = ni();
from --;
to --;
add(from, to);
}
int scc = kosarajuSCC();
int[] count = new int[scc + 1];
for (int i = 0; i < V; ++i) {
count[cmp[i]]++;
}
int ans = 0;
for (int i = 0; i <= scc; ++i) {
if (count[i] >= 2) ans ++;
}
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);
}
}
}