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.

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【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。

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【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;}