题目大意:有$n$个布尔变量 $x_1 \sim x_n$,另有$m$个需要满足的条件,每个条件的形式都是"$x_i$ 为$true/false$或$x_j$为$true/false$"。比如"$x_1$为$true$或$x_3$为$false$"、"$x_7$为$false$或$x_2$为$false$"。$2-SAT$问题的目标是给每个变量赋值使得所有条件得到满足。
题解:$2-SAT$,若$a$推出$b$,就连两条边,分别为$a -> b$和$!b -> !a$,若$a$一定为$true$,就连一条$!a -> a$($false$相同),然后$tarjan$缩点,若一个点的两个状态在同一个强连通分量中,就有矛盾,否则那一个状态先访问到就为什么状态
卡点:1.$tarjan$缩点弹出栈时条件写错
C++ Code:
#include <cstdio>
#define maxn 1000010 << 1
using namespace std;
int n, m;
int head[maxn], cnt;
struct Edge {
int to, nxt;
} e[maxn];
void add(int a, int b) {
e[++cnt] = (Edge) {b, head[a]}; head[a] = cnt;
}
int low[maxn], DFN[maxn], stack[maxn], res[maxn], tot, idx, CNT;
bool vis[maxn];
inline int min(int a, int b) {return a < b ? a : b;}
void tarjan(int rt) {
DFN[rt] = low[rt] = ++idx;
vis[stack[++tot] = rt] = true;
int v;
for (int i = head[rt]; i; i = e[i].nxt) {
v = e[i].to;
if (DFN[v]) {
if (vis[v]) low[rt] = min(low[rt], DFN[v]);
} else {
tarjan(v);
low[rt] = min(low[rt], low[v]);
}
}
if (DFN[rt] == low[rt]) {
CNT++;
do {
vis[v = stack[tot--]] = false;
res[v] = CNT;
} while (rt != v);
}
}
int main() {
scanf("%d%d", &n, &m);
for (int i = 0; i < m; i++) {
int a, b, c, d;
scanf("%d%d%d%d", &a, &b, &c, &d);
add(a << 1 | !b, c << 1 | d);
add(c << 1 | !d, a << 1 | b);
}
for (int i = 2; i <= (n << 1 | 1); i++) {
if (!DFN[i]) tarjan(i);
}
for (int i = 1; i <= n; i++) {
if (res[i << 1] == res[i << 1 | 1]) {
puts("IMPOSSIBLE");
return 0;
}
}
puts("POSSIBLE");
for (int i = 1; i <= n; i++) printf("%d ", res[i << 1] > res[i << 1 | 1]);
puts("");
return 0;
}