poj3678 Katu Puzzle 2-SAT

Katu Puzzle

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6714		Accepted: 2472

Description

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 V_i a value X_i (0 ≤ X_i ≤ 1) such that for each edge e(a, b) labeled by op and c, the following formula holds:

 X_a op X_b = 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.

Source

POJ Founder Monthly Contest – 2008.07.27, Dagger

 

先进性tarjan算法找出强连通分量，判断每个点的取值是否矛盾，矛盾即为对于题意中每个点的0，1是否存在于一个强连通分量中，矛盾输出NO，不矛盾输出1；

建图规则：

a & b = 1 等价于 !a->a, !b->b;（旨在使得a或b取值为0是产生矛盾）

a & b = 0 等价于 a->!b, b->!a;

a | b = 1  等价于 !a->b, !b->a;

a | b = 0  等价于 a->!a, b->!b;

a ^ b = 1 等价于 !a->b, !b->a, a->!b, b->!a;

a ^ b = 0 等价于 a->b, b->a, !a->!b, !b->!a;  

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int MAX = 1005 * 2;
int n,m;
int head[MAX];
struct node {
    int t,next;
}edge[4000000];
int cnt;
void add(int u, int v) {
    edge[cnt].t = v;
    edge[cnt].next = head[u];
    head[u] = cnt++;
}
int bfn[MAX];
int low[MAX];
int vis[MAX];
int s[MAX];
int sn;
void tarbfs(int u, int lay, int & scc_num) {
    vis[u] = 1;
    bfn[u] = low[u] = lay;
    s[sn++] = u;
    int i;
    for (i = head[u]; i != -1; i = edge[i].next) {
        int tmp = edge[i].t;
        if (!vis[tmp])tarbfs(tmp, ++lay, scc_num);
        if (vis[tmp] == 1)low[u] = min(low[u], low[tmp]);
    }
    if (low[u] == bfn[u]) {
        scc_num++;
        while (1) {
            sn--;
            vis[s[sn]] = 2;
            low[s[sn]] = scc_num;
            if (s[sn] == u)break;
        }
    }
}
int tarjan() {
    int scc_num = 0;
    int lay = 1;
    int i;
    sn = 0;
    memset(vis, 0, sizeof(vis));
    for (i = 0; i < n; i++) {
        if (!vis[i])
            tarbfs(i, lay, scc_num);
    }
    return scc_num;
}

int main() {
//    freopen("in.txt","r",stdin);
    int a,b,c;
    char ch[5];
    int i;
    while (scanf("%d %d",&n,&m) != EOF) {
        memset(head, -1, sizeof(head));
        n = 2 * n;
        cnt = 0;
        for (i = 0; i < m; i++) {
            scanf("%d %d %d %s",&a,&b,&c,ch);
            if (ch[0] == 'A') {
                if (c == 1) {
                    add(a<<1, a<<1|1);
                    add(b<<1, b<<1|1);
                } else {
                    add(a<<1|1, b<<1);
                    add(b<<1|1, a<<1);
                }
            } else if (ch[0] == 'O') {
                if (c == 1) {
                    add(a<<1, b<<1|1);
                    add(b<<1, a<<1|1);
                } else {
                    add(a<<1|1, a<<1);
                    add(b<<1|1, b<<1);
                }
            } else {
                if (c == 1) {
                    add(a<<1|1, b<<1);
                    add(b<<1|1, a<<1);
                    add(a<<1, b<<1|1);
                    add(b<<1, a<<1|1);
                } else {
                    add(a<<1|1, b<<1|1);
                    add(b<<1|1, a<<1|1);
                    add(a<<1, b<<1);
                    add(b<<1, a<<1);
                }
            }
        }
        tarjan();
        int flag = 0;
        for (i = 0; i < n / 2; i++) {
            if (low[i<<1] == low[i<<1|1]) {
                flag = 1;
                break;
            }
        }
        if (flag)
            printf("NO\n");
        else
            printf("YES\n");
    }
    return 0;
}