POJ 3678 Katu Puzzle（2SAT）

题意：给定一个有n个点m条边的有向图，每个边给一个运算符op（AND, OR, XOR）以及一个权值c，问是否能将每个点的值赋成Xi(0或1)后，使得每条边满足X_a op X_b = c。

分析：

1、经典的2-SAT问题。白书324页。

进行强连通分量分解后，若x和¬x在同一个强连通分量中，则无解。

2、关键在于加边：

(1)

A AND B = 1    !A->A !B->B
A AND B = 0    A->!B B->!A

(2)
A OR B = 1     !A->B !B->A
A OR B = 0     A->!A B->!B
(3)
A XOR B = 1    A->!B !B->A !A->B B->!A
A XOR B = 0    A->B B->A !A->!B !B->!A

3、对A AND B = 1的解释：

A和B都必须取1，

对于x和¬x，如果一定要取x，则连边¬x —> x。

4、a表示¬x，a+n表示x。

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
#define lowbit(x) (x & (-x))
const double eps = 1e-8;
inline int dcmp(double a, double b){
    if(fabs(a - b) < eps) return 0;
    return a > b ? 1 : -1;
}
typedef long long LL;
typedef unsigned long long ULL;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const LL LL_INF = 0x3f3f3f3f3f3f3f3f;
const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const int MOD = 1e9 + 7;
const double pi = acos(-1.0);
const int MAXN = 1000 + 10;
const int MAXT = 10000 + 10;
using namespace std;
int N;
vector<int> G[MAXN << 1];
vector<int> rG[MAXN << 1];
vector<int> vs;
bool used[MAXN << 1];
int cmp[MAXN << 1];
void addedge(int from, int to){
    G[from].push_back(to);
    rG[to].push_back(from);
}
void dfs(int v){
    used[v] = true;
    for(int i = 0; i < G[v].size(); ++i){
        if(!used[G[v][i]]) dfs(G[v][i]);
    }
    vs.push_back(v);
}
void rdfs(int v, int k){
    used[v] = true;
    cmp[v] = k;
    for(int i = 0; i < rG[v].size(); ++i){
        if(!used[rG[v][i]]) rdfs(rG[v][i], k);
    }
}
int scc(){
    memset(used, 0, sizeof used);
    vs.clear();
    for(int i = 0; i < 2 * N; ++i){
        if(!used[i]) dfs(i);
    }
    memset(used, 0, sizeof used);
    int k = 0;
    for(int i = vs.size() - 1; i >= 0; --i){
        if(!used[vs[i]]) rdfs(vs[i], k++);
    }
    return k;
}
int main(){
    int M;
    scanf("%d%d", &N, &M);
    while(M--){
        int a, b, c;
        string s;
        scanf("%d%d%d", &a, &b, &c);
        cin >> s;
        if(s == "AND"){
            if(c == 1){
                addedge(a, a + N);
                addedge(b, b + N);
            }
            else{
                addedge(a + N, b);
                addedge(b + N, a);
            }
        }
        else if(s == "OR"){
            if(c == 1){
                addedge(a, b + N);
                addedge(b, a + N);
            }
            else{
                addedge(a + N, a);
                addedge(b + N, b);
            }
        }
        else{
            if(c == 1){
                addedge(a, b + N);
                addedge(a + N, b);
                addedge(b, a + N);
                addedge(b + N, a);
            }
            else{
                addedge(a, b);
                addedge(a + N, b + N);
                addedge(b, a);
                addedge(b + N, a + N);
            }
        }
    }
    scc();
    for(int i = 0; i < N; ++i){
        if(cmp[i] == cmp[i + N]){
            printf("NO\n");
            return 0;
        }
    }
    printf("YES\n");
    return 0;
}