hdu 1824 (2-SAT)

来源：互联网发布：ubuntu只有一种分辨率编辑：程序博客网时间：2024/06/16 18:31

简单2-SAT问题,拆点，回家i，不回家i+n*3；

建好队长跟队员之间的关系，每对队员之间的关系，

判断每个队员回不回家是否矛盾

#include<stdio.h>#include<string.h>#include<stack>#define N 6060using namespace std;int low[N],n,m,dfs[N],ans,idx,belong[N],ins[N];struct Eage{int ed;Eage *next;}*E[N];void addeage(int x,int y){Eage *p=new Eage;p->ed=y;p->next=E[x];E[x]=p;}stack<int>Q;void Tarjan(int x){int v;   low[x]=dfs[x]=idx++;   ins[x]=1;   Q.push(x);   for(Eage *p=E[x];p;p=p->next)   {   v=p->ed;   if(dfs[v]==-1)   {   Tarjan(v);   low[x]=low[x]>low[v]?low[v]:low[x];   }   else if(ins[v]==1)   {   low[x]=low[x]>dfs[v]?dfs[v]:low[x];   }   }   if(low[x]==dfs[x])   {   while(1)   {   v=Q.top();   Q.pop();   ins[v]=0;   belong[v]=ans;   if(v==x)break;   }   ans++;   }}int main(){int i,a,b,c,num;while(scanf("%d%d",&n,&m)!=-1){memset(dfs,-1,sizeof(dfs));memset(ins,0,sizeof(ins));memset(E,NULL,sizeof(E));ans=idx=0;num=3*n;for(i=0;i<n;i++){scanf("%d%d%d",&a,&b,&c);addeage(a+num,b);addeage(a+num,c);addeage(b+num,a);addeage(c+num,a);}for(i=0;i<m;i++){scanf("%d%d",&a,&b);addeage(a,b+num);addeage(b,a+num);}for(i=0;i<6*n;i++){if(dfs[i]==-1)Tarjan(i);}for(i=0;i<6*n;i++){if(belong[i]==belong[i+num])break;}if(i<3*n)printf("no\n");else printf("yes\n");}return 0;}