Problem 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 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

題意：給出一 n 個點 m 條邊的有向圖，每個點只能取 0、1，現在每條邊被一個操作符 or、and、xor 以及一個值標記了，表示 a、b 按操作符運算的結果，問這個有向圖是否有可行解

思路：根據題意，該題是一個 2-SAT 問題，將操作符進行轉換，進行判定即可

對于 A[x]，可以通過連邊 <x',x> 實現，NOT A[x]，可以通過連邊 <x,x'> 來實現，對于?NOT(A[x] AND A[y]) 需要連兩條邊 <x, y'> 和 <y, x'> 來實現，對于 A[x] OR A[y] 需要連兩條邊 <x', y> 和 <y', x> 來實現

故對于 and、or、xor 三種運算有：

and 運算：

• a and b = 0 時：若 a=1，則必定滿足 b=0；若 b=1，則必定滿足 a=0，即：<a,1,b,0>、<b,1,a,0>
• a and b = 1 時：a=1 且?b=1，即：<a,0,a,1>、<b,0,b,1>

or 運算：

• a or b = 0 時：a = 0 且 b = 0，即：<a,1,a,0>、<b,1,b,0>?
• a or b = 1 時：若 a=0，則必定滿足 b=1；若?b=0，則必定滿足 a=0，即：<a,0,b,1>、<b,0,a,1>

xor 運算：

• a xor b = 0 時，有以下四種情況：
  若 a = 0，則必定滿足 b = 0，即：<a,0,b,0>
  若 b = 0，則必定滿足 a = 0，即：<b,0,a,0>
  若?a = 1，則必定滿足 b = 1，即：<a,1,b,1>
  若 b = 1，則必定滿足 a = 1，即：<b,1,a,1>
• a xor b = 1 時，有以下四種情況：
  若?a = 0，則必定滿足 b = 1，即：<a,0,b,1>
  若 b = 0，則必定滿足 a = 1，即：<b,0,a,1>
  若 a = 1，則必定滿足 b = 0，即：<a,1,b,0>
  若 b = 1，則必定滿足 a = 0，即：<b,1,a,0>

Source Program

#include<iostream> #include<cstdio> #include<cstdlib> #include<string> #include<cstring> #include<cmath> #include<ctime> #include<algorithm> #include<utility> #include<stack> #include<queue> #include<vector> #include<set> #include<map> #define PI acos(-1.0) #define E 1e-9 #define INF 0x3f3f3f3f #define LL long long const int MOD=10007; const int N=1000000+5; const int dx[]= {-1,1,0,0}; const int dy[]= {0,0,-1,1}; using namespace std;bool vis[N*2]; int Stack[N*2],top; vector<int> G[N*2]; void init(int n){memset(vis,false,sizeof(vis));for(int i=0;i<2*n;i++)G[i].clear(); } void addOrClause(int x,int xVal,int y,int yVal){x=x*2+xVal;y=y*2+yVal;G[x^1].push_back(y);G[y^1].push_back(x); } void addAndClause(int x,int xval,int y,int yval) {x=x*2+xval;y=y*2+yval;G[x].push_back(y); } bool dfs(int x){if(vis[x^1])return false;if(vis[x])return true;vis[x]=true;Stack[top++]=x;for(int i=0;i<G[x].size();i++)if(!dfs(G[x][i]))return false;return true; } bool twoSAT(int n){for(int i=0;i<2*n;i+=2){if(!vis[i] && !vis[i+1]){top=0;if(!dfs(i)){while(top>0)vis[Stack[--top]]=false;if(!dfs(i+1))return false;}}}return true; } int main(){int n,m;while(scanf("%d%d",&n,&m)!=EOF&&(n+m)){init(n);while(m--){int a,b,c;char ch[10];scanf("%d%d%d%s",&a,&b,&c,ch);if(ch[0]=='A'){//and 運算if(c==0){//a and b=0addAndClause(a,1,b,0);//若 a=1，則必定滿足 b=0addAndClause(b,1,a,0);//若 b=1，則必定滿足 a=0}else{//a and b=1addAndClause(a,0,a,1);//a=1addAndClause(b,0,b,1);//b=1}}else if(ch[0]=='O'){//or 運算if(c==0){//a or b=0addAndClause(a,1,a,0);//a=0addAndClause(b,1,b,0);//b=0}else{//a or b=1addAndClause(a,0,b,1);//若 a=0，則必定滿足 b=1addAndClause(b,0,a,1);//若 b=0，則必定滿足 a=0}}else if(ch[0]=='X'){//xor 運算if(c==0){//a xor b=0addAndClause(a,0,b,0);//若 a = 0，則必定滿足 b = 0addAndClause(b,0,a,0);//若 b = 0，則必定滿足 a = 0addAndClause(a,1,b,1);//若 a = 1，則必定滿足 b = 1addAndClause(b,1,a,1);//若 b = 1，則必定滿足 a = 1}else{//a xor b=1addAndClause(a,0,b,1);//若 a = 0，則必定滿足 b = 1addAndClause(b,0,a,1);//若 b = 0，則必定滿足 a = 1addAndClause(a,1,b,0);//若 a = 1，則必定滿足 b = 0addAndClause(b,1,a,0);//若 b = 1，則必定滿足 a = 0}}}printf("%s\n",twoSAT(n)?"YES":"NO");}return 0; }

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