luoguP4782 2-SAT问题

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);
        // a == b || c == d
        // if(a != b) c = d
        // if(c != d) a = b
        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;
}