2-sat学习笔记(补档
重学2-sat
\(2-SAT\)是什么
给定一个布尔方程,判断是否存在一组布尔变量能满足这个方程,方程的可行解被称为\(SAT\)。这个问题是\(NP-hard\)的。但是如果我们对这个问题加上一些限制,就可以在多项式时间内求解了。设一个布尔方程为:
\((a_1\lor a_2\cdots\lor a_n) \land (b_1\lor b_2\cdots\lor b_n) \cdots\)
如果一个布尔方程满足\(n<=2\),那么这就是一个\(2-SAT\)问题。
如何表示\(2-SAT\)
因为每个布尔变量只有两种取值,所以我们可以将每个布尔变量拆成两个点后建图。设将\(x\)拆成\(x\)和\(x'\)。\(x\)表示变量\(x\)值为真,\(x'\)表示变量\(x\)值为假。
如果从\(a\)的值,我们能推出\(b\)的值。那我们就在图中连一条有向边。比如一个布尔方程方程:\((a_1\lor a_2)\),我们发现如果\(a_1\)为假,那么\(a_2\)一定为真;同理如果\(a_2\)为假,那么\(a_1\)一定为真。于是我们在图中建两条有向边:\(a_1'->a_2\)和\(a_2'->a_1\)。这就是最基本的\(2-SAT\)建图方法。
如何求解\(2-SAT\)
对于一个变量\(x\),如果点\(x\)和点\(x'\)在同一个强连通分量中,那么显然这个布尔方程无解,因为此时\(x\)的值既要为真,又要为假。
如果一个布尔方程存在一组解,对于点\(x\),如果\(x\)的拓扑序在\(x'\)之后,那么答案就是\(x\)取真。反之,\(x\)就取假。因为这样能够避免冲突,构造出一组可行解
求解细节
先在图中求一遍强连通分量。可以用\(kosaraj\)或者\(tarjan\)。注意\(kosaraju\)中强连通分量编号的顺序就是该图强连通分量的拓扑序。而在\(tarjan\)则是强联通分量编号顺序则和该图强连通分量拓扑序相反。
代码
代码中强联通分量用\(kosaraju\)求解
graph::kosaraju();
for (int i = 1; i <= n; ++i) {
if (bel[i] == bel[i + n]) {
puts("IMPOSSIBLE");
return 0;
}
}
puts("POSSIBLE");
for (int i = 1; i <= n; ++i) {
printf("%d ", bel[i] > bel[i + n]);
}
例题
洛谷P4782
题意
求解\(2-SAT\)问题
题解
对每种情况分别连边即可。设当前两个变量分别为\(A\)和\(B\)
\(\begin{cases} A'->B, B'->A & A真B真\\ A'->B', B->A & A真B假 \\ A->B, B'->A' & A假B真 \\ A->B', B->A' & A假B假 \end{cases}\)
然后按照上面讲的方法跑\(kosaraju\)求解即可。
// Author: 23forever
#include <bits/stdc++.h>
#define pb push_back
#define pii pair<int, int>
#define mp make_pair
#define fi first
#define se second
typedef long long LL;
const int MAXN = 2000000;
using namespace std;
inline int read() {
int x = 0, w = 1;
char c = ' ';
while (!isdigit(c)) {
c = getchar();
if (c == '-') w = -1;
}
while (isdigit(c)) {
x = (x << 1) + (x << 3) + (c ^ 48);
c = getchar();
}
return x * w;
}
int n, m, bel[MAXN + 5];
namespace graph {
const int MAXM = 4000000;
struct Edge {
int to, nxt;
Edge() {}
Edge(int _to, int _nxt) : to(_to), nxt(_nxt) {}
} edge[MAXM + 5], redge[MAXM + 5];
int tot, head[MAXN + 5], rtot, rhead[MAXN + 5];
inline void addEdge(int u, int v) {
edge[tot] = Edge(v, head[u]), head[u] = tot++;
redge[rtot] = Edge(u, rhead[v]), rhead[v] = rtot++;
}
inline void init() {
tot = 0, memset(head, -1, sizeof(head));
rtot = 0, memset(rhead, -1, sizeof(rhead));
}
vector < int > vec;
bool vis[MAXN + 5];
void dfs1(int u) {
vis[u] = true;
for (int i = head[u]; ~i; i = edge[i].nxt) {
int v = edge[i].to;
if (!vis[v]) dfs1(v);
}
vec.pb(u);
}
void dfs2(int u, int cnt) {
vis[u] = true;
bel[u] = cnt;
for (int i = rhead[u]; ~i; i = redge[i].nxt) {
int v = redge[i].to;
if (!vis[v]) dfs2(v, cnt);
}
}
void kosaraju() {
for (int i = 1; i <= 2 * n; ++i) {
if (!vis[i]) dfs1(i);
}
memset(vis, false, sizeof(vis));
int cnt = 0;
for (int i = vec.size() - 1; ~i; --i) {
if (!vis[vec[i]]) dfs2(vec[i], ++cnt);
}
}
}
void init() {
graph::init();
n = read(), m = read();
for (int t = 1; t <= m; ++t) {
int i = read(), a = read(), j = read(), b = read();
if (a && b) {
graph::addEdge(i + n, j);
graph::addEdge(j + n, i);
} else if (a && !b) {
graph::addEdge(i + n, j + n);
graph::addEdge(j, i);
} else if (!a && b) {
graph::addEdge(i, j);
graph::addEdge(j + n, i + n);
} else {
graph::addEdge(i, j + n);
graph::addEdge(j, i + n);
}
}
}
int main() {
#ifdef forever23
freopen("test.in", "r", stdin);
//freopen("test.out", "w", stdout);
#endif
init();
graph::kosaraju();
for (int i = 1; i <= n; ++i) {
if (bel[i] == bel[i + n]) {
puts("IMPOSSIBLE");
return 0;
}
}
puts("POSSIBLE");
for (int i = 1; i <= n; ++i) {
printf("%d ", bel[i] > bel[i + n]);
}
return 0;
}
洛谷P4171
题意
每个变量有两种取值,所以问题就是求解一个\(2-SAT\)
题解
建图和上一题一模一样:
\(\begin{cases} A'->B, B'->A & A:h,B:h\\ A'->B', B->A & A:h,B:m \\ A->B, B'->A' & A:m,B:h \\ A->B', B->A' & A:m,B:m \end{cases}\)
只要看有没有\(x\)和\(x'\)在同一个强联通分量即可。
代码
// Author: 23forever
#include <bits/stdc++.h>
typedef long long LL;
const int MAXN = 200;
using namespace std;
inline int read() {
int x = 0, w = 1;
char c = ' ';
while (!isdigit(c)) {
c = getchar();
if (c == '-') w = -1;
}
while (isdigit(c)) {
x = (x << 1) + (x << 3) + (c ^ 48);
c = getchar();
}
return x * w;
}
int n, m, bel[MAXN + 5], scc_num;
namespace graph {
const int MAXM = 2000;
struct Edge {
int to, nxt;
Edge() {}
Edge(int _to, int _nxt) : to(_to), nxt(_nxt) {}
} edge[MAXM + 5];
int tot, head[MAXN + 5];
inline void addEdge(int u, int v) {
edge[tot] = Edge(v, head[u]), head[u] = tot++;
}
inline void init() {
tot = 0, memset(head, -1, sizeof(head));
}
int dfn[MAXN + 5], low[MAXN + 5], idx;
bool in_sta[MAXN + 5];
stack < int > sta;
void dfs(int u) {
dfn[u] = low[u] = ++idx;
sta.push(u);
in_sta[u] = true;
for (int i = head[u]; ~i; i = edge[i].nxt) {
int v = edge[i].to;
if (!dfn[v]) {
dfs(v);
low[u] = min(low[u], low[v]);
} else if (in_sta[v]) {
low[u] = min(low[u], dfn[v]);
}
}
if (dfn[u] == low[u]) {
++scc_num;
int k;
do {
k = sta.top(), bel[k] = scc_num;
in_sta[k] = false, sta.pop();
} while (k != u);
}
}
void tarjan() {
memset(dfn, 0, sizeof(dfn));
for (int i = 1; i <= 2 * n; ++i) {
if (!dfn[i]) dfs(i);
}
}
}
void init() {
graph::init();
n = read(), m = read();
for (int i = 1; i <= m; ++i) {
char c1 = getchar();
while (!isalpha(c1)) c1 = getchar();
int x = read();
char c2 = getchar();
while (!isalpha(c2)) c2 = getchar();
int y = read();
if (c1 == 'h' && c2 == 'h') {
graph::addEdge(y + n, x);
graph::addEdge(x + n, y);
} else if (c1 == 'h' && c2 == 'm') {
graph::addEdge(x + n, y + n);
graph::addEdge(y, x);
} else if (c1 == 'm' && c2 == 'h') {
graph::addEdge(x, y);
graph::addEdge(y + n, x + n);
} else {
graph::addEdge(x, y + n);
graph::addEdge(y, x + n);
}
}
}
int main() {
#ifdef forever23
freopen("test.in", "r", stdin);
//freopen("test.out", "w", stdout);
#endif
int t = read();
while (t--) {
init();
graph::tarjan();
bool f = false;
for (int i = 1; i <= n; ++i) {
if (bel[i] == bel[i + n]) {
puts("BAD");
f = true;
break;
}
}
if (!f) puts("GOOD");
}
return 0;
}
总结
求解\(2-SAT\)问题其实是通过巧妙的建图方法来解决。这种能推出就拉边的思想十分常见。在并查集相关题目中也能见到。值得积累。