https://www.luogu.org/problemnew/show/P4782
2-SAT模板,输出方案只需判断 (a) 和 (a + n) 两个点所在的 scc 编号大小就可以了
#include <bits/stdc++.h>
using namespace std;
template <typename T>
inline void read(T &f) {
f = 0; T fu = 1; char c = getchar();
while (c < '0' || c > '9') {if (c == '-') fu = -1; c = getchar();}
while (c >= '0' && c <= '9') {f = (f << 3) + (f << 1) + (c & 15); c = getchar();}
f *= fu;
}
const int N = 2000000 + 10;
struct Edge {
int u, v, next;
}G[N << 1];
int head[N], col[N], low[N], dfn[N], st[N], inst[N];
int n, m, tot, cnt, Index, len;
inline void addedge(int u, int v) {
G[++tot] = (Edge) {u, v, head[u]}, head[u] = tot;
}
void tarjan(int u) {
low[u] = dfn[u] = ++Index;
st[++len] = u; inst[u] = 1;
for(int i = head[u]; i; i = G[i].next) {
int v = G[i].v;
if(!dfn[v]) {
tarjan(v);
low[u] = min(low[u], low[v]);
} else if(inst[v]) low[u] = min(low[u], dfn[v]);
}
if(low[u] == dfn[u]) {
cnt++;
while(st[len + 1] != u) {
int tmp = st[len--];
col[tmp] = cnt;
inst[tmp] = 0;
}
}
}
int main() {
read(n); read(m);
for(int i = 1; i <= m; i++) {
int a, b, c, d;
read(a); read(b); read(c); read(d);
if(b == 0) {
if(d == 0) addedge(a + n, c), addedge(c + n, a);
else addedge(a + n, c + n), addedge(c, a);
} else {
if(d == 0) addedge(a, c), addedge(c + n, a + n);
else addedge(a, c + n), addedge(c, a + n);
}
}
for(int i = 1; i <= (n << 1); i++) if(!dfn[i]) tarjan(i);
for(int i = 1; i <= n; i++) if(col[i] == col[i + n]) {puts("IMPOSSIBLE"); return 0;}
puts("POSSIBLE");
for(int i = 1; i <= n; i++) if(col[i] > col[i + n]) printf("1 "); else printf("0 ");
return 0;
}