[2-SAT] poj 3678 Katu Puzzle
题目链接：
http://poj.org/problem?id=3678
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Katu Puzzle
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:
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. 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.
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 2200
int 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;
}