POJ3678 Katu Puzzle

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 9238		Accepted: 3436

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

只需要判断可行性，不用求值。

2-sat妥妥的。

模拟运算符模式连边，跑一边tarjan就行。

  1 #include<iostream>
  2 #include<cstdio>
  3 #include<cstring>
  4 #include<algorithm>
  5 #include<cmath>
  6 using namespace std;
  7 const int mxn=12000;
  8 //bas
  9 int n,m;
 10 //edge
 11 struct edge{
 12     int to;
 13     int next;
 14 }e[mxn*100];
 15 int hd[mxn],cnt=0;
 16 void add_edge(int u,int v){
 17     e[++cnt].next=hd[u];e[cnt].to=v;hd[u]=cnt;
 18 }
 19 
 20 //tarjan
 21 int vis[mxn];
 22 int dfn[mxn],low[mxn];
 23 int st[mxn],top;
 24 bool inst[mxn];
 25 int dtime=0;
 26 int belone[mxn],tot;
 27 void tarjan(int s){
 28     low[s]=dfn[s]=++dtime;
 29     st[++top]=s;
 30     inst[s]=1;
 31     for(int i=hd[s];i;i=e[i].next){
 32         int v=e[i].to;
 33         if(!dfn[v]){
 34             tarjan(v);
 35             low[s]=min(low[s],low[v]);
 36         }
 37         else if(inst[v]){
 38             low[s]=min(low[s],dfn[v]);
 39         }
 40     }
 41     int v;
 42     if(low[s]==dfn[s]){
 43         cnt++;
 44         do{    
 45             v=st[top--];
 46             inst[v]=0;
 47             belone[v]=cnt;
 48             
 49         }while(v!=s);
 50     }
 51     return;
 52 }
 53 void Build(int a,int b,int c,char op){
 54     switch(op){
 55         case 'A':{
 56             if(c==1){
 57                 add_edge(a+n,a);//原点表示选1, +n点表示0； 
 58                 add_edge(b+n,b);
 59                 add_edge(a,b);
 60                 add_edge(b,a);
 61                 
 62             }
 63             else{
 64                 add_edge(a,b+n);
 65                 add_edge(b,a+n);                
 66             }
 67             break;
 68         }
 69         case 'O':{
 70             if(c==1){
 71                 add_edge(a+n,b);
 72                 add_edge(b+n,a);
 73             }
 74             else{
 75                 add_edge(a,a+n); 
 76                 add_edge(b,b+n);
 77                 add_edge(a+n,b+n);
 78                 add_edge(b+n,a+n);
 79             }
 80             break;
 81         }
 82         case 'X':{
 83             if(c==1){
 84                 add_edge(a,b+n);
 85                 add_edge(b,a+n);
 86                 add_edge(a+n,b);
 87                 add_edge(b+n,a);
 88             }
 89             else{
 90                 add_edge(a,b);
 91                 add_edge(b,a);
 92                 add_edge(b+n,a+n);
 93                 add_edge(a+n,b+n);
 94             }
 95             break;
 96         }
 97         
 98     }
 99 }
100 int main(){
101     scanf("%d%d",&n,&m);
102     int i,j;
103     int a,b,c;
104     char op[10];
105     for(i=1;i<=m;i++){
106         scanf("%d%d%d%s",&a,&b,&c,op);
107         a++;b++;
108         Build(a,b,c,op[0]);
109     }
110     for(i=1;i<=2*n;i++)if(!dfn[i])tarjan(i);
111     for(i=1;i<=n;i++){
112         if(belone[i]==belone[i+n] && belone[i]!=0){
113             printf("NO
");
114             return 0;
115         }
116     }
117     printf("YES
");
118     return 0;
119 }