POJ 3678 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:

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.

【题目分析】

其实暴搜加上适当的剪枝可以0ms的，但是熟悉一下2set，建图的方式非常巧妙。

【代码】

#include <cstdio>
#include <vector>
using namespace std;
typedef long long ll;
int dfn[2001],vis[2001],low[2001],sta[2001*10],bel[2001],tem,cnt,top;
vector<int>e[2001];
void tarjan(int u)
{
dfn[u]=low[u]=++tem;
vis[u]=true;
sta[++top]=u;
int v,i,l=e[u].size();
for(i=0;i<l;i++)
{
v=e[u][i];
if(!dfn[v])
{
tarjan(v);
low[u]=min(low[u],low[v]);
}
else if(vis[v]&&dfn[v]<low[u]) low[u]=dfn[v];
}
if(dfn[u]==low[u])
{
cnt++;
do
{
v=sta[top--];
vis[v]=false;
bel[v]=cnt;
}while(v!=u);
}
}
bool twoSAT(int n)
{
for(int i=0;i<2*n;i++)
if(!dfn[i]) tarjan(i);
for(int i=0;i<n;i++)
if(bel[2*i]==bel[2*i+1]) return false;
return true;
}
int main()
{
int n,m,i,a,b,c;
char s[5];
scanf("%d%d",&n,&m);
for(i=0;i<m;i++)
{
scanf("%d%d%d%s",&a,&b,&c,s);
if(s[0]=='A')
{
if(c) e[2*a+1].push_back(2*a),e[2*b+1].push_back(2*b);
else e[2*a].push_back(2*b+1),e[2*b].push_back(2*a+1);
}
else if(s[0]=='O')
{
if(c) e[2*a+1].push_back(2*b),e[2*b+1].push_back(2*a);
else e[2*a].push_back(2*a+1),e[2*b].push_back(2*b+1);
}
}
if(twoSAT(n)) puts("YES");
else puts("NO");
return 0;
}