挑战程序竞赛系列（79）：4.3 2-SAT（3）
传送门：POJ 2723: Get Luffy Out
题意：
题目意思有点坑，实际上给出每一对钥匙，如（0，3），如果选择了钥匙0，那么后续的门只能用钥匙0开，而不能再选择3，意味着每对钥匙是可以开多扇门的。
思路： 二分 2-SAT，每一扇门有两把锁，一扇门能推出一对矛盾关系，如OJ上提供的数据：
代码语言：javascript复制3 6
0 3
1 2
4 5
0 1
0 2
4 1
4 2
3 5
2 2
0 0
第一扇门为：(0,1)
那么对应的钥匙串有(0,3)和(1,2)
开任意一把锁都能通过该层，如下：
钥匙串A 钥匙串B 门
0 1 通过
0 2 通过
3 1 通过
3 2 未通过
所以此处矛盾关系为3和2，即钥匙串1和钥匙串2不能同时选择3和2。
有了矛盾关系就可以根据2-SAT模型去解了。
二分：
显然，矛盾关系越多（每扇门一个约束条件），符合条件的钥匙串选择就越少，符合单调性，加快搜索。代码如下：
代码语言：javascript复制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/P2723.txt";
public static void main(String[] args) throws IOException {
new Main().run();
}
int[][] doors;
int[] keys;
int N, M;
class SCC {
int V;
List<Integer>[] g;
List<Integer>[] rg;
List<Integer> po;
boolean[] used;
int[] cmp;
SCC(int V){
this.V = V;
g = new List[V];
rg = new List[V];
po = new ArrayList<Integer>();
used = new boolean[V];
cmp = new int[V];
for (int i = 0; i < V; i) g[i] = new ArrayList<Integer>();
for (int i = 0; i < V; i) rg[i] = new ArrayList<Integer>();
}
void add(int from, int to) {
g[from].add(to);
rg[to].add(from);
}
void dfs(int v) {
used[v] = true;
for (int u : g[v]) {
if (!used[u]) dfs(u);
}
po.add(v);
}
void rdfs(int v, int k) {
used[v] = true;
cmp[v] = k;
for (int u : rg[v]) {
if (!used[u]) rdfs(u, 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) {
int v = po.get(i);
if (!used[v]) rdfs(v, k );
}
return k;
}
}
void read() {
while (true) {
N = ni();
M = ni();
if (N M == 0) break;
keys = new int[2 * N];
for (int i = 0; i < N; i) {
int A = ni();
int B = ni();
keys[A] = B;
keys[B] = A;
}
doors = new int[M][2];
for (int i = 0; i < M; i) {
doors[i] = new int[] {ni(), ni()};
}
solve();
}
}
void solve() {
int lf = 0, rt = M;
while (lf < rt) {
int mid = lf (rt - lf 1) / 2;
if (!valid(mid)) {
rt = mid - 1;
}
else {
lf = mid;
}
}
out.println(lf);
}
boolean valid(int m) {
SCC scc = new SCC(2 * N);
for (int i = 0; i < m; i) {
scc.add(keys[doors[i][0]], doors[i][1]);
scc.add(keys[doors[i][1]], doors[i][0]);
}
scc.kosarajuSCC();
for (int i = 0; i < scc.V; i) {
if (scc.cmp[i] == scc.cmp[keys[i]]) {
return false;
}
}
return true;
}
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);
}
}
static class D{
public static void pp(int[][] board, int row, int col) {
StringBuilder sb = new StringBuilder();
for (int i = 0; i < row; i) {
for (int j = 0; j < col; j) {
sb.append(board[i][j] (j 1 == col ? "n" : " "));
}
}
System.out.println(sb.toString());
}
public static void pp(char[][] board, int row, int col) {
StringBuilder sb = new StringBuilder();
for (int i = 0; i < row; i) {
for (int j = 0; j < col; j) {
sb.append(board[i][j] (j 1 == col ? "n" : " "));
}
}
System.out.println(sb.toString());
}
}
static class ArrayUtils {
public static void fill(int[][] f, int value) {
for (int i = 0; i < f.length; i) {
Arrays.fill(f[i], value);
}
}
public static void fill(int[][][] f, int value) {
for (int i = 0; i < f.length; i) {
fill(f[i], value);
}
}
public static void fill(int[][][][] f, int value) {
for (int i = 0; i < f.length; i) {
fill(f[i], value);
}
}
}
static class Num{
public static <K> void inc(Map<K, Integer> mem, K k) {
if (!mem.containsKey(k)) mem.put(k, 0);
mem.put(k, mem.get(k) 1);
}
}
}
