给大家推荐一个关于2-SAT讲的不错的博客（自我认为）
2-SAT

#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
const int N = 4e6+10;

int h[N],e[N],ne[N],idx;
int low[N], dfn[N], c[N], stack[N];
vector<int> scc[N];
int n, m, num, top, cnt;
bool ins[N];

void add(int a,int b) {
    e[++idx] = b, ne[idx] = h[a], h[a] = idx;
}

void tarjan(int x) {
    dfn[x] = low[x] = ++num;
    ins[x] = 1;
    stack[++top] = x;
    for(int i = h[x]; i; i = ne[i]) {
        int y = e[i];
        if(!dfn[y]) {
            tarjan(y);
            low[x] = min(low[x],low[y]);
        } else if(ins[y])
            low[x] = min(low[x], dfn[y]);
    }
    if(low[x] == dfn[x]) {
        int z;
        cnt++;
        do {
            z = stack[top--];
            c[z] = cnt;
            ins[z] = false;
            scc[cnt].push_back(z);
        } while(x!=z);
    }
}

int main() {
    scanf("%d",&n);
    scanf("%d",&m);
    for(int i = 1; i <= m; i++) {
        int a,x,b,y;
        scanf("%d%d%d%d",&a,&x,&b,&y);
        //"如果第a个数为0或第b个数为0"至少满足其一  a=1则b=0  b=1则a=0 
        if(x == 0 && y == 0)  add(a + n,b), add(b + n,a);
        //如果第a个数为0或第b个数为1"至少满足其一   a=1则b=1  b=0则a=0 
        if(x == 0 && y == 1)  add(a + n,b + n), add(b, a);
        //"如果第a个数为1或第b个数为0"至少满足其一  a=0则b=0  b=1则a=1 
        if(x == 1 && y == 0)  add(a, b), add(b + n, a + n);
        // "如果第a个数为1或第b个数为1"至少满足其一 a=0则b=1  b=0则a=1 
        if(x == 1 && y == 1)  add(a, b + n), add(b, a + n);
    }

    for(int i = 1; i <= 2 * n; i++)
        if(!dfn[i]) tarjan(i);

    for(int x = 1; x <= n; x++) {
        if(c[x] == c[x+n]) {
            printf("IMPOSSIBLE\n");
            return 0;
        }
    }

    printf("POSSIBLE\n");
    for(int i = 1; i <= n; i++) {
        //强联通分量编号越小 -> 拓扑序越大 -> 越优 
        if(c[i] > c[i+n]) printf("1 ");
        else printf("0 ");
    }
    return 0;
}