2SAT问题解题报告
2SAT是P问题，能在多项式时间内解决。这里对方法之一进行总结。
测试数据格式如下：
1000000
-383037 -585864
575603 -494722
-628477 805085
-502855 685868
923336 -606921
-653863 424389
333542 -656925
-809061 874253
即范围为1~1000000，“-”号表示取逆。比如-383037 -585864表示~383037U~585864，575603 -494722表示575603U~494722
判断是否存在一种取法，能够满足上述所有的clause。
解法是用图论中的强连通子图。具体来说，首先构造一个有向图。每个数字和它的逆都是一个单独的节点。对于每个条目，如AUB，则在图中加入两条边：~A->B和~B->A。即A如果为负，则B为正；反之，如果B为负，则A为正。这两条边和AUB是等效的，都制约了A和B的取值。构造好有向图后就构建强连通子图。一个强连通子图中的变量相互制约，必须同时成立。
所以，如果有一对变量A和~A在同一个强连通子图中，则该2SAT无法满足。因为变量A和~A是无法同时成立的。
public class TwoSAT {
public static void main(String[] args)
{
In in = new In(args[0]);
int N = in.readInt();
int V = 2 * N;
Digraph digraph = new Digraph(V);
while (!in.isEmpty())
{
int v = in.readInt();
int w = in.readInt();
int negv = 2 * Math.abs(v) - 1, posv = 2 * Math.abs(v) - 2;
int negw = 2 * Math.abs(w) - 1, posw = 2 * Math.abs(w) - 2;
digraph.addEdge(v > 0 ? negv : posv, w > 0 ? posw : negw);
digraph.addEdge(w > 0 ? negw : posw, v > 0 ? posv : negv);
}
StdOut.println("input done.");
KosarajuSharirSCC scc = new KosarajuSharirSCC(digraph);
boolean satisfied = true;
for (int i = 0; i < N; ++i)
{
if (scc.stronglyConnected(2 * i, 2 * i + 1))
{
satisfied = false;
break;
}
}
if (satisfied)
{
StdOut.println(true);
} else
{
StdOut.println(false);
}
}
}
其中用到了Algos4中的求强连通子图的方法，过程为求逆，加两遍dfs。
/*************************************************************************
* Compilation: javac KosarajuSharirSCC.java
* Execution: java KosarajuSharirSCC filename.txt
* Dependencies: Digraph.java TransitiveClosure.java StdOut.java In.java
* Data files: http://algs4.cs.princeton.edu/42directed/tinyDG.txt
*
* Compute the strongly-connected components of a digraph using the
* Kosaraju-Sharir algorithm.
*
* Runs in O(E + V) time.
*
* % java KosarajuSCC tinyDG.txt
* 5 components
* 1
* 0 2 3 4 5
* 9 10 11 12
* 6
* 7 8
*
* % java KosarajuSharirSCC mediumDG.txt
* 10 components
* 21
* 2 5 6 8 9 11 12 13 15 16 18 19 22 23 25 26 28 29 30 31 32 33 34 35 37 38 39 40 42 43 44 46 47 48 49
* 14
* 3 4 17 20 24 27 36
* 41
* 7
* 45
* 1
* 0
* 10
*
*************************************************************************/
public class KosarajuSharirSCC {
private boolean[] marked; // marked[v] = has vertex v been visited?
private int[] id; // id[v] = id of strong component containing v
private int count; // number of strongly-connected components
public KosarajuSharirSCC(Digraph G) {
// compute reverse postorder of reverse graph
DepthFirstOrder dfs = new DepthFirstOrder(G.reverse());
// run DFS on G, using reverse postorder to guide calculation
marked = new boolean[G.V()];
id = new int[G.V()];
for (int v : dfs.reversePost()) {
if (!marked[v]) {
dfs(G, v);
count++;
}
}
// check that id[] gives strong components
assert check(G);
}
// DFS on graph G
private void dfs(Digraph G, int v) {
marked[v] = true;
id[v] = count;
for (int w : G.adj(v)) {
if (!marked[w]) dfs(G, w);
}
}
// return the number of strongly connected components
public int count() { return count; }
// are v and w strongly connected?
public boolean stronglyConnected(int v, int w) {
return id[v] == id[w];
}
// id of strong component containing v
public int id(int v) {
return id[v];
}
// does the id[] array contain the strongly connected components?
private boolean check(Digraph G) {
TransitiveClosure tc = new TransitiveClosure(G);
for (int v = 0; v < G.V(); v++) {
for (int w = 0; w < G.V(); w++) {
if (stronglyConnected(v, w) != (tc.reachable(v, w) && tc.reachable(w, v)))
return false;
}
}
return true;
}
public static void main(String[] args) {
In in = new In(args[0]);
Digraph G = new Digraph(in);
KosarajuSharirSCC scc = new KosarajuSharirSCC(G);
// number of connected components
int M = scc.count();
StdOut.println(M + " components");
// compute list of vertices in each strong component
Queue<Integer>[] components = (Queue<Integer>[]) new Queue[M];
for (int i = 0; i < M; i++) {
components[i] = new Queue<Integer>();
}
for (int v = 0; v < G.V(); v++) {
components[scc.id(v)].enqueue(v);
}
// print results
for (int i = 0; i < M; i++) {
for (int v : components[i]) {
StdOut.print(v + " ");
}
StdOut.println();
}
}
}