HOJ 1917 Peaceful Commission

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03年的论文题

《伍昱--由对称性解2-SAT问题》

突然发现我以前写的2-SAT原来是O(nm)的（lrj书上并没有提到复杂度）

然后就学会了O(m+n)的2-SAT，不过有个很坑爹的地方就是这个算法不能求字典序最小的方案。

#include<iostream>#include<cstdio>#include<cstring>#include<stack>#include<queue>#include<vector>using namespace std;struct Edge{int to,next,v;}e[200005];int head[16005],cnt;void ins(int u,int v){e[++cnt]=(Edge){v,head[u]};head[u]=cnt;}int pr[16005],n,m;int sccno[16005],scc_cnt,dfs_clock,pre[16005],lowlink[16005];stack<int>s;vector<int>g[16005];void tarjan(int u){lowlink[u]=pre[u]=++dfs_clock;s.push(u);for(int i=head[u];i;i=e[i].next){int v=e[i].to;if(!pre[v]){tarjan(v);lowlink[u]=min(lowlink[u],lowlink[v]);}else if(!sccno[v])lowlink[u]=min(lowlink[u],pre[v]);}if(lowlink[u]==pre[u]){scc_cnt++;while(true){int x=s.top();s.pop();sccno[x]=scc_cnt;if(x==u)break;}}}void find_scc(){for(int i=1;i<=2*n;i++)if(!pre[i])tarjan(i);}int du[16005];void change(){for(int i=1;i<=scc_cnt;i++)g[i].clear();for(int u=1;u<=2*n;u++)for(int i=head[u];i;i=e[i].next){int v=e[i].to;if(sccno[u]!=sccno[v])g[sccno[v]].push_back(sccno[u]),du[sccno[u]]++;}}int color[16005];void toposort(){queue<int>q;for(int i=1;i<=scc_cnt;i++)if(!du[i])q.push(i);while(!q.empty()){int u=q.front();q.pop();if(!color[u])color[u]=1,color[pr[u]]=2;for(int i=0;i<g[u].size();i++){int v=g[u][i];du[v]--;if(!du[v])q.push(v);}}}void init(){memset(head,0,sizeof(head));cnt=0;memset(sccno,0,sizeof(sccno));memset(lowlink,0,sizeof(lowlink));memset(pre,0,sizeof(pre));dfs_clock=scc_cnt=0;while(!s.empty())s.pop();memset(du,0,sizeof(du));memset(color,0,sizeof(color));}void solve(){find_scc();for(int i=1;i<=2*n;i+=2)if(sccno[i]==sccno[i+1]){printf("NIE\n");return;}else pr[sccno[i]]=sccno[i+1],pr[sccno[i+1]]=sccno[i];change();toposort();for(int i=1;i<=2*n;i+=2)if(color[sccno[i]]==1)printf("%d\n",i);else printf("%d\n",i+1);}int main(){//freopen("a.in","r",stdin);//freopen("a.out","w",stdout);while(~scanf("%d%d",&n,&m)){init();int u,v;for(int i=1;i<=m;i++){scanf("%d%d",&u,&v);ins(u,((v-1)^1)+1);ins(v,((u-1)^1)+1);}solve();}return 0;}

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