poj3648：Wedding——题解（配2-SAT简易讲解）

#include<stack>
#include<cstdio>
#include<cstring>
#include<vector>
#include<iostream>
#include<algorithm>
using namespace std;const int N=1005;
const int M=100005;
struct node{
    int to;
    int nxt;
}edge[M*4];
int head[N*2],dfn[N*2],low[N*2],to[N*2];
int n,m,t,l,cnt;
bool instack[N*2];
stack<int>q;
inline void add(int u,int v){
    cnt++;
    edge[cnt].to=v;
    edge[cnt].nxt=head[u];
    head[u]=cnt;
    return;
}
inline int neg(int x){
    if(x<=n)return x+n;
    return x-n;
}
void tarjan(int u){
    t++;
    dfn[u]=t;
    low[u]=t;
    q.push(u);
    instack[u]=1;
    for(int i=head[u];i!=0;i=edge[i].nxt){
    int v=edge[i].to;
    if(!dfn[v]){
        tarjan(v);
        low[u]=min(low[u],low[v]);
    }else if(instack[v]){
        low[u]=min(low[u],dfn[v]);
    }
    }
    if(low[u]==dfn[u]){
    int v;
    l++;
    do{
        v=q.top();
        q.pop();
        instack[v]=0;
        to[v]=l;
    }while(v!=u);
    }
    return;
}
inline void clr(){
    cnt=0;t=0;l=0;
    memset(dfn,0,sizeof(dfn));
    memset(head,0,sizeof(head));
    return;
}
int main(){
    while(scanf("%d%d",&n,&m)!=EOF&&(n>0||m>0)){
    clr();
    for(int i=1;i<=m;i++){
        int u,v;char a,b;
        scanf("%d%c %d%c",&u,&a,&v,&b);
        u++;v++;
        if(a==‘h‘)u+=n;
        if(b==‘h‘)v+=n;
        add(u,neg(v));
        add(v,neg(u));
    }
    add(1,n+1);
    for(int i=1;i<=n*2;i++){
        if(!dfn[i])tarjan(i);
    }
    bool flag=1;
    for(int i=1;i<=n;i++){
        if(to[i]==to[i+n]){
        flag=0;
        break;
        }
    }
    if(!flag){
        printf("bad luck\n");
        continue;
    }
    for(int i=2;i<=n;i++){
        printf("%d",i-1);
        if(to[i]>to[i+n])printf("w ");
        else printf("h ");
    }
    printf("\n");
    }
    return 0;
}