poj 3678

      题目意思是，给出一系列的二元布尔运算，问所给变量是否存在可解的情况。这些布尔运算有与、或和异或运算。

       经典的2-sat问题。由于变量之间隐含着“冲突”关系，可以通过构造2-sat模型来求解。具体的建图规则为：

       若 a and b ==1 , !a->a , !b -> b

            a and b ==0 , a->!b , b->!a

            a or b ==1 , !a->b , !b->a

            a or b ==0 , a->!a , b->!b

            a xor b ==1 , a->!b,!b->a,!a->b,b->!a

            a xor b ==0 , a->b,b->a,!a->!b,!b->!a

 

#include<cstdio>#include<cstring>#include<iostream>using namespace std;const int M=1000007;const int N=1020;struct node{int v;int next;}edge[4*M];int head1[2*N],head2[2*N],num;int id[2*N],ord[2*N];bool vis[2*N];bool flag;int n,m,cnt;void init(){for(int i=0;i<=2*n+1;i++){head1[i]=-1;head2[i]=-1;id[i]=0;}num=0;flag=true;}void addege1(int u,int v){edge[num].v=v;edge[num].next=head1[u];head1[u]=num++;}void addege2(int u,int v){edge[num].v=v;edge[num].next=head2[u];head2[u]=num++;}void build(){int  a,b,w;char s[10];for(int i=0;i<m;i++){scanf("%d%d%d%s",&a,&b,&w,s);if(strcmp(s,"AND")==0){if(w==1){addege1(a*2+1,a*2);addege1(b*2+1,b*2);addege2(a*2,a*2+1);addege2(b*2,b*2+1);}else{addege1(a*2,b*2+1);addege1(b*2,a*2+1);addege2(b*2+1,a*2);addege2(a*2+1,b*2);}}else if(strcmp(s,"OR")==0){if(w==1){addege1(a*2+1,b*2);addege1(b*2+1,a*2);addege2(b*2,a*2+1);addege2(a*2,b*2+1);}else{addege1(a*2,a*2+1);addege1(b*2,b*2+1);addege2(a*2+1,a*2);addege2(b*2+1,b*2);}}else if(strcmp(s,"XOR")==0){if(w==1){addege1(a*2,b*2+1);addege1(b*2+1,a*2);addege1(a*2+1,b*2);addege1(b*2,a*2+1);addege2(b*2+1,a*2);addege2(a*2,b*2+1);addege2(b*2,a*2+1);addege2(a*2+1,b*2);}else{addege1(a*2,b*2);addege1(b*2,a*2);addege1(a*2+1,b*2+1);addege1(b*2+1,a*2+1);addege2(b*2,a*2);addege2(a*2,b*2);addege2(b*2+1,a*2+1);addege2(a*2+1,b*2+1);}}}}void dfs1(int u){vis[u]=1;for(int i=head1[u];i!=-1;i=edge[i].next){int v=edge[i].v;if(!vis[v]) dfs1(v);}ord[cnt++]=u;}void dfs2(int u){vis[u]=1;id[u]=cnt;for(int i=head2[u];i!=-1;i=edge[i].next){int v=edge[i].v;if(!vis[v]) dfs2(v);}}void kosaraju(){cnt=0;fill(vis,vis+2*n,false);for(int i=0;i<2*n;i++){if(!vis[i]) dfs1(i);}cnt=0;fill(vis,vis+2*n,false);for(int i=2*n-1;i>=0;i--){if(!vis[ord[i]]){ dfs2(ord[i]);++cnt;}}--cnt;for(int i=0;i<n;i++){if(id[2*i]==id[2*i+1]){flag=false;return ;}}}int main(){while(scanf("%d%d",&n,&m)==2){init();build();kosaraju();if(flag) puts("YES");else puts("NO");}return 0;}