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  • luoguP4782 [模板]2-SAT问题

    https://www.luogu.org/problemnew/show/P4782

    2-SAT模板,输出方案只需判断 (a)(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;
    } 
    
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  • 原文地址:https://www.cnblogs.com/LJC00118/p/9509013.html
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