题目链接：

题目

建图：

1.新娘向新郎连边，表示选取的时候选取新娘对面的。

2.

有奸情的人比如是 xh yh那么连边(x,y+n)以及(y,x+n)

如果是xh yw 那么连边(x,y) 以及 (y+n,x)

如果是xw yh 那么连边(x+n,y+n)以及(y,x)

如果是xw yw那么连边(x+n,y)以及(y,x+n)

其中  (x表示丈夫 x <= n ，x+n表示妻子)

3.跑2-sat，输出的时候注意输出新娘一边的。

因为我们选取的时候是按照新郎一边的选择的，所以说输出的时候要取反再输出。

代码：

#include <vector>
#include <cstring>
#include <cstdio>
#include <queue>
using namespace std;
const int maxn = 1000;
int head[maxn];
int DFN[maxn],LOW[maxn],stk[maxn],visit[maxn],belong[maxn];
vector<int> scc[maxn];
int tot,idx,cnt,sccnum;
int n,color[maxn],degree[maxn],pos[maxn];
struct Es{ int v; int next; int cost; 
}Es[maxn<<1];  
vector<int> bkG[maxn];
void init(){ sccnum = tot = idx = cnt = 0; memset(head,-1,sizeof(head)); memset(DFN,0,sizeof(DFN)); memset(LOW,0,sizeof(LOW)); memset(visit,0,sizeof(visit)); memset(color,0,sizeof(color)); memset(degree,0,sizeof(degree));memset(pos,0,sizeof(pos));for(int i = 0;i < maxn;i++) scc[i].clear();for(int i = 0;i < maxn;++i) bkG[i].clear();
}
inline void add_edge(int i,int j,int cost = 1){   Es[cnt].v = j; Es[cnt].cost = cost; Es[cnt].next = head[i]; head[i] = cnt++; 
}   
void tarjan(int x)
{DFN[x]=LOW[x]=++tot;stk[++idx]=x;visit[x]=1;for(int i=head[x];i!=-1;i=Es[i].next){if(!DFN[Es[i].v]) {tarjan(Es[i].v);LOW[x] = min(LOW[x],LOW[Es[i].v]);}else if(visit[Es[i].v ]){  LOW[x] = min(LOW[x],DFN[Es[i].v]);}}if(LOW[x]==DFN[x]){++sccnum;do{int item = stk[idx];belong[item] = sccnum;scc[sccnum].push_back(item);visit[item]=0;idx--;}while(x!=stk[idx+1]);}return ;
}
bool check(){for(int i = 1;i <= n;i++){if(belong[i] == belong[i+n]) return false;}return true;
}
bool solve(){for(int i = 1;i <= 2*n;++i )if(!DFN[i]) tarjan(i);if(!check()) return false;for(int i = 1;i <= n;++i){if(!pos[belong[i]]){pos[belong[i]] = belong[i+n];pos[belong[i+n]] = belong[i];}}for(int i = 1;i <= 2*n;++i){for(int e = head[i];e != -1;e = Es[e].next){int v = Es[e].v;if(belong[i] != belong[v]){degree[belong[i]]++;bkG[belong[v]].push_back(belong[i]);}}}queue<int> que;for(int i = 1;i <= sccnum;++i){if(!degree[i])que.push(i);}while(!que.empty()){int u = que.front();que.pop();if(!color[u]){color[u] = 1;color[pos[u]] = 2;}for(int i = 0;i < bkG[u].size();++i){int v = bkG[u][i];degree[v]--;if(!degree[v])que.push(v);}}//在这里准备输出答案//为1表示选中，为2表示为选中/*for(int i = 1;i <= n;++i){if(color[belong[i]] == 1) {//Yes}else{//No}} */return true;
}
int main(){int m;while(~scanf("%d%d",&n,&m) && n + m){init();//n--;for(int i = 0;i < m;++i){int a = 0,b = 0;char ta,tb;scanf("%d%c%d%c",&a,&ta,&b,&tb);a++;b++;add_edge(ta == 'h'?a:a+n,tb == 'h'?b+n:b);add_edge(tb == 'h'?b:b+n,ta == 'h'?a+n:a);}add_edge(n+1,1);if(solve()){for(int i = 2;i <= n;i++){if(color[belong[i]] == 1){printf("%d%c ",i-1,'w');}else{printf("%d%c ",i-1,'h');}}puts("");}else{puts("bad luck");}}} 
/*
3 2
1w 2w
1w 2h*/