poj 3678 Katu Puzzle 2-SAT 建图入门

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

X ₀ = 1,  X ₁ = 1,  X ₂ = 0,  X ₃ = 1.

 　　根据上面直接建图

 

　　

  1 #include <iostream>
  2 #include <cstdio>
  3 #include <cstring>
  4 #include <vector>
  5 #include <string>
  6 #include <algorithm>
  7 #include <queue>
  8 #include <stack>
  9 
 10 using namespace std;
 11 const int maxn = 1e5 + 10;
 12 const int  mod = 1e9 + 7 ;
 13 const int INF = 0x7ffffff;
 14 struct node {
 15     int v, next;
 16 } edge[maxn];
 17 int head[maxn], dfn[maxn], low[maxn];
 18 int s[maxn], belong[maxn], instack[maxn];
 19 int tot, cnt, top, flag, n, m;
 20 void init() {
 21     tot = cnt = top = flag = 0;
 22     memset(s, 0, sizeof(s));
 23     memset(head, -1, sizeof(head));
 24     memset(dfn, 0, sizeof(dfn));
 25     memset(instack, 0, sizeof(instack));
 26 }
 27 void add(int u, int v ) {
 28     edge[tot].v = v;
 29     edge[tot].next = head[u];
 30     head[u] = tot++;
 31 }
 32 void tarjan(int v) {
 33     dfn[v] = low[v] = ++flag;
 34     instack[v] = 1;
 35     s[top++] = v;
 36     for (int i = head[v] ; ~i ; i = edge[i].next ) {
 37         int j = edge[i].v;
 38         if (!dfn[j]) {
 39             tarjan(j);
 40             low[v] = min(low[v], low[j]);
 41         } else if (instack[j]) low[v] = min(low[v], dfn[j]);
 42     }
 43     if (dfn[v] == low[v]) {
 44         cnt++;
 45         int t;
 46         do {
 47             t = s[--top];
 48             instack[t] = 0;
 49             belong[t] = cnt;
 50         } while(t != v) ;
 51     }
 52 }
 53 int check() {
 54     for (int i = 0 ; i < n ; i++)
 55         if (belong[2 * i] == belong[2 * i + 1]) return 0;
 56     return 1;
 57 }
 58 int main() {
 59     while(scanf("%d%d", &n, &m) != EOF) {
 60         if (n == 0 && m == 0) break;
 61         init();
 62         char op[10];
 63         int x, y, c;
 64         for (int i = 0 ; i < m ; i++) {
 65             scanf("%d%d%d%s", &x, &y, &c, op);
 66             if (op[0] == 'A') {
 67                 if (c) {
 68                     add(2 * x + 1, 2 * x);
 69                     add(2 * y + 1, 2 * y);
 70                 } else {
 71                     add(2 * x, 2 * y + 1);
 72                     add(2 * y, 2 * x + 1);
 73                 }
 74             }
 75             if (op[0] == 'O') {
 76                 if (c) {
 77                     add(2 * x + 1, 2 * y);
 78                     add(2 * y + 1, 2 * x);
 79                 } else {
 80                     add(2 * x, 2 * x + 1);
 81                     add(2 * y, 2 * y + 1);
 82                 }
 83             }
 84             if (op[0] == 'X') {
 85                 if (c) {
 86                     add(2 * x, 2 * y + 1);
 87                     add(2 * x + 1, 2 * y);
 88                     add(2 * y, 2 * x + 1);
 89                     add(2 * y + 1, 2 * x);
 90                 } else {
 91                     add(2 * x + 1, 2 * y + 1);
 92                     add(2 * x, 2 * y);
 93                     add(2 * y + 1, 2 * x + 1);
 94                     add(2 * y, 2 * x);
 95                 }
 96             }
 97 
 98         }
 99         for (int i = 0 ; i < 2 * n ; i++)
100             if (!dfn[i]) tarjan(i);
101         if (check()) printf("YES\n");
102         else printf("NO\n");
103     }
104     return 0;
105 }