题目链接:http://poj.org/problem?id=3648‘
题意:一对情人举行婚礼,有n对夫妇参加,别人对着坐在一个长桌子的两边,新娘和新郎坐在最后面,新娘只能看见坐在他对面的人。现在,n对夫妇中有两两通奸了,新娘不希望看到这种情况,先给出通奸的人,求是否存在可行的排位方法。
有两种建立2sat模型的方法,第一种是对每个人的坐的方向来建立,0和1分别表示坐在左边和右边。但是可以不考虑坐方向,只考虑他们能不能坐在同一边,需要考虑新郎的一边,因为新娘那边得不出限制条件。
1 //STATUS:C++_AC_16MS_180KB 2 #include <functional> 3 #include <algorithm> 4 #include <iostream> 5 //#include <ext/rope> 6 #include <fstream> 7 #include <sstream> 8 #include <iomanip> 9 #include <numeric> 10 #include <cstring> 11 #include <cassert> 12 #include <cstdio> 13 #include <string> 14 #include <vector> 15 #include <bitset> 16 #include <queue> 17 #include <stack> 18 #include <cmath> 19 #include <ctime> 20 #include <list> 21 #include <set> 22 #include <map> 23 using namespace std; 24 //using namespace __gnu_cxx; 25 //define 26 #define pii pair<int,int> 27 #define mem(a,b) memset(a,b,sizeof(a)) 28 #define lson l,mid,rt<<1 29 #define rson mid+1,r,rt<<1|1 30 #define PI acos(-1.0) 31 //typedef 32 typedef long long LL; 33 typedef unsigned long long ULL; 34 //const 35 const int N=110; 36 const int INF=0x3f3f3f3f; 37 const int MOD=5000,STA=100010; 38 const LL LNF=1LL<<60; 39 const double EPS=1e-8; 40 const double OO=1e15; 41 const int dx[4]={-1,0,1,0}; 42 const int dy[4]={0,1,0,-1}; 43 const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31}; 44 //Daily Use ... 45 inline int sign(double x){return (x>EPS)-(x<-EPS);} 46 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;} 47 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;} 48 template<class T> inline T lcm(T a,T b,T d){return a/d*b;} 49 template<class T> inline T Min(T a,T b){return a<b?a:b;} 50 template<class T> inline T Max(T a,T b){return a>b?a:b;} 51 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);} 52 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);} 53 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));} 54 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));} 55 //End 56 57 int first[N*2],next[N*N*2],vis[N*2],S[N*2]; 58 int n,m,mt,cnt; 59 60 struct Edge{ 61 int u,v; 62 }e[N*N*2]; 63 64 void adde(int a,int b) 65 { 66 e[mt].u=a,e[mt].v=b; 67 next[mt]=first[a];first[a]=mt++; 68 } 69 70 int dfs(int u) 71 { 72 if(vis[u^1])return 0; 73 if(vis[u])return 1; 74 int i; 75 vis[u]=1; 76 S[cnt++]=u; 77 for(i=first[u];i!=-1;i=next[i]){ 78 if(!dfs(e[i].v))return 0; 79 } 80 return 1; 81 } 82 83 int Twosat() 84 { 85 int i,j; 86 for(i=0;i<n;i+=2){ 87 if(vis[i] || vis[i^1])continue; 88 cnt=0; 89 if(!dfs(i)){ 90 while(cnt)vis[S[--cnt]]=0; 91 if(!dfs(i^1))return 0; 92 } 93 } 94 return 1; 95 } 96 97 int main() 98 { 99 // freopen("in.txt","r",stdin); 100 int i,j,a,b,x,y,flag; 101 char c1,c2; 102 while(~scanf("%d%d",&n,&m) && (n||m)) 103 { 104 n<<=1; 105 mem(first,-1); 106 mem(vis,0);mt=0; 107 adde(1,0); 108 for(i=0;i<m;i++){ 109 scanf("%d%c%d%c",&a,&c1,&b,&c2); 110 x=(a<<1)+(c1=='h'?0:1); 111 y=(b<<1)+(c2=='h'?0:1); 112 // printf("%d %d %d %d\n",x,y,x^1,y^1); 113 adde(x,y^1); 114 adde(y,x^1); 115 } 116 117 if(Twosat()){ 118 // for(i=0;i<n;i+=2) 119 // printf("%d %d\n",vis[i],vis[i+1]); 120 121 printf("%d%c",1,vis[2]==vis[1]?'h':'w'); 122 for(i=4;i<n;i+=2){ 123 printf(" %d%c",i/2,vis[i]==vis[1]?'h':'w'); 124 } 125 } 126 else printf("bad luck"); 127 putchar('\n'); 128 } 129 return 0; 130 }