【算法训练】POJ - 3678 Katu Puzzle （2-SAT）

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题意

有$n$个布尔变量，$m$个条件，求是否有可行解。

题解

2-SAT
如果必须$u$，那么$u'->u$。
tarjan缩点判一哈儿原变量和反变量是否在一个scc中好了。

代码

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef double db;
typedef long long ll;
typedef unsigned long long ull;
const int nmax = 1e6+7;
const int INF = 0x3f3f3f3f;
const ll LINF = 0x3f3f3f3f3f3f3f3f;
const ull p = 67;
const ull MOD = 1610612741;
int head[nmax], tot;
struct edge {
    int to, nxt;
}e[nmax<<1];
void add_edge(int u, int v) {
    e[tot].to = v;
    e[tot].nxt = head[u];
    head[u] = tot++;
}
int dfn[nmax], low[nmax], color[nmax], ss[nmax], st, dfs_clock, scc;
bool instack[nmax];
void tarjan(int u) {
    dfn[u] = low[u] = ++ dfs_clock;
    ss[st++] = u;
    instack[u] = true;
    for(int i = head[u]; i != -1; i = e[i].nxt) {
        int v = e[i].to;
        if(!dfn[v]) {
            tarjan(v);
            low[u] = min(low[u], low[v]);
        } else if(instack[v]) {
            low[u] = min(low[u], dfn[v]);
        }
    }
    if(low[u] == dfn[u]) {
        scc ++;
        int tmp;
        for(;;) {
            tmp = ss[--st];
            instack[tmp] = false;
            color[tmp] = scc;
            if(tmp == u)
                break;
        }
    }
}
void init() {
    memset(head, -1, sizeof head);
    tot = scc = dfs_clock = st = 0;
    memset(dfn, 0 ,sizeof dfn);
    memset(low, 0, sizeof low);
    memset(color, 0, sizeof color);
}
int n , m;
int main() {
    while(scanf("%d %d", &n, &m) != EOF){
        init();
        int u, v, res;
        char op[10];
        for(int i = 1; i <= m; ++i) {
            scanf("%d %d %d %s", &u, &v, &res, op);
            u ++;
            v ++;
            if(op[0] == 'A') {
                if(res == 1) {
                    add_edge(u + n, u);
                    add_edge(v + n, v);
                } else {
                    add_edge(u, v + n);
                    add_edge(v, u + n);
                }
            } else if(op[0] == 'O') {
                if(res == 1) {
                    add_edge(u + n, v);
                    add_edge(v + n, u);
                } else {
                    add_edge(u, u + n);
                    add_edge(v, v + n);
                }
            } else if(op[0] == 'X') {
                if(res == 1) {
                    add_edge(u, v + n);
                    add_edge(u + n, v);
                    add_edge(v, u + n);
                    add_edge(v + n, u);
                } else {
                    add_edge(u, v);
                    add_edge(v, u);
                    add_edge(u + n, v + n);
                    add_edge(v + n, u + n);
                }
            }
        }
        for(int i = 1; i  <= 2 * n; ++i) {
            if(!dfn[i])
                tarjan(i);
        }
        bool isok = true;
        for(int i = 1; i <= n; ++i) {
            if(color[i]  == color[i + n]){
                isok = false;
                break;
            }
        }
        if(isok) puts("YES");
        else puts("NO");
    }
    return 0;
}