[2-Sat] poj 3678 Katu puzzle

题目链接：

http://poj.org/problem?id=3678

Katu Puzzle Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 7888   Accepted: 2888 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 40 1 1 AND1 2 1 OR3 2 0 AND3 0 0 XOR Sample Output YES Hint X0 = 1, X1 = 1, X2 = 0, X3 = 1. Source POJ Founder Monthly Contest – 2008.07.27, Dagger

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题目意思：

给一幅有向图，每条边代表连接的两个节点的一种运算，并且告诉运算后的值，问能否有一种节点取值，使得各边都满足边的运算。

解题思路：

2-SAT

i代表1 ~i带表0

当i&j==1时，建边  ~i->i     ~j->j 

当i&j==0时，建边   i->~j    j->~i  ~i->~j   ~j->~i

当i|j==1时，建边    ~i->j     ~j->i   

当i|j==0时，建边   i->~i    j->~j

当i^j==1时，建边  i->~j  ~i->j   j->~i  ~j->i

当i^j==0时，建边  i->j  ~i->~j   j->i  ~j->~i

代码：

//#include<CSpreadSheet.h>#include<iostream>#include<cmath>#include<cstdio>#include<sstream>#include<cstdlib>#include<string>#include<string.h>#include<cstring>#include<algorithm>#include<vector>#include<map>#include<set>#include<stack>#include<list>#include<queue>#include<ctime>#include<bitset>#include<cmath>#define eps 1e-6#define INF 0x3f3f3f3f#define PI acos(-1.0)#define ll __int64#define LL long long#define lson l,m,(rt<<1)#define rson m+1,r,(rt<<1)|1#define M 1000000007//#pragma comment(linker, "/STACK:1024000000,1024000000")using namespace std;#define Maxn 2200int n,m;vector<vector<int> >myv;int low[Maxn],dfn[Maxn],sc,bc,dep;int in[Maxn],sta[Maxn];bool iss[Maxn];void tarjan(int cur){    int ne;    low[cur]=dfn[cur]=++dep;    sta[++sc]=cur;    iss[cur]=true;    for(int i=0;i<myv[cur].size();i++)    {        ne=myv[cur][i];        if(!dfn[ne])        {            tarjan(ne);            low[cur]=min(low[cur],low[ne]);        }        else if(iss[ne]&&dfn[ne]<low[cur])            low[cur]=dfn[ne];    }    if(low[cur]==dfn[cur])    {        ++bc;        do        {            ne=sta[sc--];            in[ne]=bc;            iss[ne]=false;        }while(ne!=cur);    }}void solve(){    sc=bc=dep=0;    memset(dfn,0,sizeof(dfn));    memset(iss,false,sizeof(iss));    for(int i=0;i<2*n;i++)        if(!dfn[i])            tarjan(i);}int main(){    //freopen("in.txt","r",stdin);   //freopen("out.txt","w",stdout);   while(~scanf("%d%d",&n,&m))   {       myv.clear();       myv.resize(2*n+10);       for(int i=1;i<=m;i++)       {           int a,b,c;           char ss[10];           scanf("%d%d%d%s",&a,&b,&c,ss);           if(*ss=='A')           {               if(c==1)               {                   myv[2*a].push_back(2*a+1);                   myv[2*a+1].push_back(2*b+1);                   myv[2*b].push_back(2*b+1);                   myv[2*b+1].push_back(2*a+1);               }               else               {                   myv[2*a+1].push_back(2*b);                   myv[2*b+1].push_back(2*a);               }           }           else if(*ss=='O')           {               if(c==1)               {                   myv[2*a].push_back(2*b+1);                   myv[2*b].push_back(2*a+1);               }               else               {                   myv[2*a+1].push_back(2*a);                   myv[2*a].push_back(2*b);                   myv[2*b].push_back(2*a);                   myv[2*b+1].push_back(2*b);               }           }           else           {               if(c==1)               {                   myv[2*a].push_back(2*b+1);                   myv[2*a+1].push_back(2*b);                   myv[2*b].push_back(2*a+1);                   myv[2*b+1].push_back(2*a);               }               else               {                   myv[2*a].push_back(2*b);                   myv[2*a+1].push_back(2*b+1);                   myv[2*b].push_back(2*a);                   myv[2*b+1].push_back(2*a+1);               }           }       }       solve();       bool ans=true;       for(int i=0;i<n;i++)       {           if(in[2*i]==in[2*i+1])           {               ans=false;               break;           }       }       if(ans)            printf("YES\n");       else            printf("NO\n");   }    return 0;}

[2-SAT] poj 3678 Katu Puzzle