POJ-3678 Katu Puzzle 2sat
题目链接：http://poj.org/problem?id=3678
分别对and，or，xor推出相对应的逻辑关系：
逻辑关系 1 0
A and B A'->A,B'->B A->B',B->A'
A or B A'->B',B'->A A->A',B->B'
A xor B A'->B,B'->A,A->B',B->A' A->B,A'->B'
1 //STATUS:C++_AC_96MS_472KB 2 #include <functional> 3 #include <algorithm> 4 #include <iostream> 5 //#include <ext/rope> 6 #include <fstream> 7 #include <sstream> 8 #include <iomanip> 9 #include <numeric> 10 #include <cstring> 11 #include <cassert> 12 #include <cstdio> 13 #include <string> 14 #include <vector> 15 #include <bitset> 16 #include <queue> 17 #include <stack> 18 #include <cmath> 19 #include <ctime> 20 #include <list> 21 #include <set> 22 #include <map> 23 using namespace std; 24 //using namespace __gnu_cxx; 25 //define 26 #define pii pair<int,int> 27 #define mem(a,b) memset(a,b,sizeof(a)) 28 #define lson l,mid,rt<<1 29 #define rson mid+1,r,rt<<1|1 30 #define PI acos(-1.0) 31 //typedef 32 typedef long long LL; 33 typedef unsigned long long ULL; 34 //const 35 const int N=1010; 36 const int INF=0x3f3f3f3f; 37 const int MOD=5000,STA=100010; 38 const LL LNF=1LL<<60; 39 const double EPS=1e-8; 40 const double OO=1e15; 41 const int dx[4]={-1,0,1,0}; 42 const int dy[4]={0,1,0,-1}; 43 const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31}; 44 //Daily Use ... 45 inline int sign(double x){return (x>EPS)-(x<-EPS);} 46 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;} 47 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;} 48 template<class T> inline T lcm(T a,T b,T d){return a/d*b;} 49 template<class T> inline T Min(T a,T b){return a<b?a:b;} 50 template<class T> inline T Max(T a,T b){return a>b?a:b;} 51 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);} 52 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);} 53 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));} 54 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));} 55 //End 56 57 int first[N*2],next[N*N*2],vis[N*2],S[N*2]; 58 int n,m,mt,cnt; 59 60 struct Edge{ 61 int u,v; 62 }e[N*N*2]; 63 64 void adde(int a,int b) 65 { 66 e[mt].u=a,e[mt].v=b; 67 next[mt]=first[a];first[a]=mt++; 68 } 69 70 int dfs(int u) 71 { 72 if(vis[u^1])return 0; 73 if(vis[u])return 1; 74 int i; 75 vis[u]=1; 76 S[cnt++]=u; 77 for(i=first[u];i!=-1;i=next[i]){ 78 if(!dfs(e[i].v))return 0; 79 } 80 return 1; 81 } 82 83 int Twosat() 84 { 85 int i,j; 86 mem(vis,0); 87 for(i=0;i<n;i+=2){ 88 if(vis[i] || vis[i^1])continue; 89 cnt=0; 90 if(!dfs(i)){ 91 while(cnt)vis[S[--cnt]]=0; 92 if(!dfs(i^1))return 0; 93 } 94 } 95 return 1; 96 } 97 98 int main() 99 { 100 // freopen("in.txt","r",stdin); 101 int i,j,a,b,c; 102 char op[5]; 103 while(~scanf("%d%d",&n,&m) && (n||m)) 104 { 105 n<<=1; 106 mem(first,-1);mt=0; 107 while(m--){ 108 scanf("%d%d%d%s",&a,&b,&c,op); 109 a<<=1,b<<=1; 110 if(op[0]=='A'){ 111 if(c){ 112 adde(a^1,a); 113 adde(b^1,a); 114 } 115 else { 116 adde(a,b^1); 117 adde(b,a^1); 118 } 119 } 120 else if(op[0]=='O'){ 121 if(c){ 122 adde(a^1,b); 123 adde(b^1,a); 124 } 125 else { 126 adde(a,a^1); 127 adde(b,b^1); 128 } 129 } 130 else { 131 if(c){ 132 adde(a^1,b); 133 adde(b^1,a); 134 adde(a,b^1); 135 adde(b,a^1); 136 } 137 else { 138 adde(a,b); 139 adde(a^1,b^1); 140 } 141 } 142 } 143 144 printf("%s\n",Twosat()?"YES":"NO"); 145 } 146 return 0; 147 }
