| Time Limit: 1000MS | Memory Limit: 65536K | |||
| Total Submissions: 6919 | Accepted: 2127 | Special Judge | ||
Description
Up to thirty couples will attend a wedding feast, at which they will be seated on either side of a long table. The bride and groom sit at one end, opposite each other, and the bride wears an elaborate headdress that keeps her from seeing people on the same side as her. It is considered bad luck to have a husband and wife seated on the same side of the table. Additionally, there are several pairs of people conducting adulterous relationships (both different-sex and same-sex relationships are possible), and it is bad luck for the bride to see both members of such a pair. Your job is to arrange people at the table so as to avoid any bad luck.
Input
The input consists of a number of test cases, followed by a line containing 0 0. Each test case gives n, the number of couples, followed by the number of adulterous pairs, followed by the pairs, in the form "4h 2w" (husband from couple 4, wife from couple 2), or "10w 4w", or "3h 1h". Couples are numbered from 0 to n - 1 with the bride and groom being 0w and 0h.
Output
For each case, output a single line containing a list of the people that should be seated on the same side as the bride. If there are several solutions, any one will do. If there is no solution, output a line containing "bad luck".
Sample Input
10 6 3h 7h 5w 3w 7h 6w 8w 3w 7h 3w 2w 5h 0 0
Sample Output
1h 2h 3w 4h 5h 6h 7h 8h 9h----------
TwoSAT.add_con(0,0);设置0始终为0
------------
/** head-file **/
#include <iostream>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <list>
#include <set>
#include <map>
#include <algorithm>
/** define-for **/
#define REP(i, n) for (int i=0;i<int(n);++i)
#define FOR(i, a, b) for (int i=int(a);i<int(b);++i)
#define DWN(i, b, a) for (int i=int(b-1);i>=int(a);--i)
#define REP_1(i, n) for (int i=1;i<=int(n);++i)
#define FOR_1(i, a, b) for (int i=int(a);i<=int(b);++i)
#define DWN_1(i, b, a) for (int i=int(b);i>=int(a);--i)
#define REP_N(i, n) for (i=0;i<int(n);++i)
#define FOR_N(i, a, b) for (i=int(a);i<int(b);++i)
#define DWN_N(i, b, a) for (i=int(b-1);i>=int(a);--i)
#define REP_1_N(i, n) for (i=1;i<=int(n);++i)
#define FOR_1_N(i, a, b) for (i=int(a);i<=int(b);++i)
#define DWN_1_N(i, b, a) for (i=int(b);i>=int(a);--i)
/** define-useful **/
#define clr(x,a) memset(x,a,sizeof(x))
#define sz(x) int(x.size())
#define see(x) cerr<<#x<<" "<<x<<endl
#define se(x) cerr<<" "<<x
#define pb push_back
#define mp make_pair
/** test **/
#define Display(A, n, m) {
REP(i, n){
REP(j, m) cout << A[i][j] << " ";
cout << endl;
}
}
#define Display_1(A, n, m) {
REP_1(i, n){
REP_1(j, m) cout << A[i][j] << " ";
cout << endl;
}
}
using namespace std;
/** typedef **/
typedef long long LL;
/** Add - On **/
const int direct4[4][2]={ {0,1},{1,0},{0,-1},{-1,0} };
const int direct8[8][2]={ {0,1},{1,0},{0,-1},{-1,0},{1,1},{1,-1},{-1,1},{-1,-1} };
const int direct3[6][3]={ {1,0,0},{0,1,0},{0,0,1},{-1,0,0},{0,-1,0},{0,0,-1} };
const int MOD = 1000000007;
const int INF = 0x3f3f3f3f;
const long long INFF = 1LL << 60;
const double EPS = 1e-9;
const double OO = 1e15;
const double PI = acos(-1.0); //M_PI;
const int maxn=2111;
const int maxm=2111111;
int n,m;
struct EDGENODE{
int to;
int next;
};
struct TWO_SAT{
int head[maxn*2];
EDGENODE edges[maxm*2];
int edge;
int n;
bool mark[maxn*2];
int S[maxn*2],c;
void init(int n){
this->n=n;
clr(mark,0);
clr(head,-1);
edge=0;
}
void addedge(int u,int v){
edges[edge].to=v,edges[edge].next=head[u],head[u]=edge++;
}
// x = xval or y = yval
void add_clause(int x,int xval,int y,int yval){
x=x*2+xval;
y=y*2+yval;
addedge(x^1,y);
addedge(y^1,x);
}
void add_con(int x,int xval){
x=x*2+xval;
addedge(x^1,x);
}
bool dfs(int x){
if (mark[x^1]) return false;
if (mark[x]) return true;
mark[x]=true;
S[c++]=x;
for (int i=head[x];i!=-1;i=edges[i].next)
if (!dfs(edges[i].to)) return false;
return true;
}
bool solve(){
for (int i=0;i<n*2;i+=2)
if (!mark[i]&&!mark[i+1]){
c=0;
if (!dfs(i)){
//if (i==0) return false;
while (c>0) mark[S[--c]]=false;
if (!dfs(i+1)) return false;
}
}
return true;
}
}TwoSAT;
int main()
{
while (~scanf("%d%d",&n,&m))
{
if (n==0&&m==0) break;
TwoSAT.init(n);
REP(i,m)
{
int x,y;
char c1,c2;
cin>>x>>c1>>y>>c2;
if (c1=='h'&&c2=='h') TwoSAT.add_clause(x,1,y,1);
if (c1=='w'&&c2=='w') TwoSAT.add_clause(x,0,y,0);
if (c1=='h'&&c2=='w') TwoSAT.add_clause(x,1,y,0);
if (c1=='w'&&c2=='h') TwoSAT.add_clause(x,0,y,1);
}
//TwoSAT.addedge(1,0);
TwoSAT.add_con(0,0);
if (TwoSAT.solve()){
for (int i=1;i<n;i++){
if (TwoSAT.mark[i*2]) printf("%dw",i);
else printf("%dh",i);
if (i!=n-1) printf(" ");
}
puts("");
}else{
puts("bad luck");
}
}
return 0;
}