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 }

 

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