POJ 3678 2-SAT简单题

POJ 3678 2-SAT

题目链接

Katu Puzzle is presented as a directed graph G(V, E) with each edge e(a, b) labeled by a boolean operator op (one of AND, OR, XOR) and an integer c (0 ≤ c ≤ 1). One Katu is solvable if one can find each vertex Vi a value Xi (0 ≤ Xi ≤ 1) such that for each edge e(a, b) labeled by op and c, the following formula holds:

Xa op Xb = c
The calculating rules are:
AND 0 1
0 0 0
1 0 1
OR 0 1
0 0 1
1 1 1
XOR 0 1
0 0 1
1 1 0

Given a Katu Puzzle, your task is to determine whether it is solvable.

Input

The first line contains two integers N (1 ≤ N ≤ 1000) and M,(0 ≤ M ≤ 1,000,000) indicating the number of vertices and edges.
The following M lines contain three integers a (0 ≤ a < N), b(0 ≤ b < N), c and an operator op each, describing the edges.

Output

Output a line containing "YES" or "NO".

Sample Input

4 4

0 1 1 AND

1 2 1 OR

3 2 0 AND

3 0 0 XOR

Sample Output

YES

Hint
X0 = 1, X1 = 1, X2 = 0, X3 = 1.

题意

n个点,m个关系,每个关系给a,b,c三个数字，和一个字符串(or,and,xor)表示
a opt b == c,然后问你是否有一种取值满足所有关系

思路

2-SAT,xi 表示变量i取0, xi+n表示i取1.把关系变成有向图，如果i, i + n在同一强连通分量中那么显然关系错误。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <stack>

using namespace std;
const int MAXN = 2e3 + 7;
const int MAXM = 2e6 + 7;

struct Node {
    int to, w, next;
} edge[MAXM * 2];

int first[MAXN], sign, dfn[MAXN], low[MAXN], ins[MAXN], n, m, idx, tot, color[MAXN];

stack<int>st;

inline void init() {
    memset(first, -1, sizeof(first));
    memset(dfn, 0, sizeof(dfn));
    memset(low, 0, sizeof(low));
    memset(ins, 0, sizeof(ins));
    memset(color, 0, sizeof(color));
    sign = idx = tot = 0;
    while(!st.empty()) {
        st.pop();
    }
}

inline void add_edge(int u, int v, int w) {
    edge[sign].to = v;
    edge[sign].w = w;
    edge[sign].next = first[u];
    first[u] = sign++;
}

void tarjan(int x) {
    low[x] = dfn[x] = ++idx;
    st.push(x), ins[x] = 1;
    for(int i = first[x]; ~i; i = edge[i].next) {
        int to = edge[i].to;
        if(!dfn[to]) {
            tarjan(to);
            low[x] = min(low[x], low[to]);
        } else if(ins[to]) {
            low[x] = min(low[x], dfn[to]);
        }
    }
    int top;
    if(dfn[x] == low[x]) {
        tot++;
        do {
            top = st.top();
            st.pop();
            ins[top] = 0;
            color[top] = tot;
        } while(top != x);
    }
}

void TwoSAT() {
    for(int i = 1; i <= n; i++ ) {
        if(!dfn[i]) {
            tarjan(i);
        }
    }
    bool ok = 1;
    for(int i = 1; i <= n; i++ ) {
        if(color[i] == color[i + n]) {
            ok = 0;
            break;
        }
    }
    puts(ok ? "YES" : "NO");
}

int main() {
    int a, b, c;
    char opt[15];
    while(~scanf("%d %d", &n, &m)) {
        init();
        for(int i = 1; i <= m; i++ ) {
            scanf("%d %d %d %s", &a, &b, &c, opt);
            a++, b++;
            if(opt[0] == 'A') {
                if(c == 1) { /// A and B == 1;
                    add_edge(a, a + n, 1), add_edge(b, b + n, 1);
                    add_edge(a + n, b + n, 1), add_edge(b + n, a + n, 1);
                } else {     /// A and B == 0;
                    add_edge(a + n, b, 1);
                    add_edge(b + n, a, 1);
                }
            } else if(opt[0] == 'O') {
                if(c == 1) { /// A or B == 1;
                    add_edge(a, b + n, 1), add_edge(b, a + n, 1);
                } else {     /// A or B == 0;
                    add_edge(a, b, 1), add_edge(b, a, 1);
                    add_edge(a + n, a, 1), add_edge(b + n, b, 1);
                }
            } else if(opt[0] == 'X') {
                if(c == 1) { /// A xor B == 1;
                    add_edge(a, b + n, 1), add_edge(b, a + n, 1);
                    add_edge(a + n, b, 1), add_edge(b + n, a, 1);
                } else {     /// A xor B == 0;
                    add_edge(a, b, 1), add_edge(b, a, 1);
                    add_edge(a + n, b + n, 1), add_edge(b + n, a + n, 1);
                }
            }
        }
        TwoSAT();
    }

    return 0;
}