2-sat->poj 3648 Wedding

Wedding

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7406		Accepted: 2254		Special Judge

DescriptionUp 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 sameside 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 itis 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.InputThe 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 fromcouple 2), or "10w 4w", or "3h 1h". Couples are numbered from 0 to n - 1 with the bride and groom being 0w and 0h.OutputFor 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

构图求解：

新娘编号0， 从n - 1 对夫妇中选择n - 1个不相容的坐在新娘的对面。加边add（0， n），新娘必不选。

#include <iostream>#include <algorithm>#include <string>#include <string.h>#include <stack>#include <queue>using namespace std;#define MAXN 200 + 10struct Edge{int u, v, next;}edge[200 * MAXN], edgetwo[200 * MAXN];int head[MAXN], e, etwo, cnt, ind, n, m;int dfn[MAXN], low[MAXN], belong[MAXN], cf[MAXN], col[MAXN];int indegree[MAXN], headtwo[MAXN];bool ins[MAXN];void add(int u, int v){edge[e].u = u;edge[e].v = v;edge[e].next = head[u];head[u] = e++;}void insert(int u, int v){edgetwo[etwo].v = v;edgetwo[etwo].next = headtwo[u];headtwo[u] = etwo++;}queue <int> q;stack <int> s;void init(){e = etwo = ind = cnt = 0;while (!q.empty()){q.pop();}while (!s.empty()){s.pop();}memset(head, -1, sizeof(head));memset(headtwo, -1, sizeof(headtwo));memset(dfn, 0, sizeof(dfn));memset(low, 0, sizeof(low));memset(indegree, 0, sizeof(indegree));memset(cf, 0, sizeof(cf));memset(col, 0, sizeof(col));memset(belong, 0, sizeof(belong));memset(ins, false, sizeof(ins));}void tarjan(int u){dfn[u] = low[u] = ++ind;ins[u] = true;s.push(u);for (int i = head[u]; i != -1; i = edge[i].next){int v = edge[i].v;if (!dfn[v]){tarjan(v);low[u] = min(low[u], low[v]);}else if (ins[v]){low[u] = min(low[u], dfn[v]);}}if (low[u] == dfn[u]){cnt++;int x;do{x = s.top();s.pop();ins[x] = false;belong[x] = cnt;}while (x != u);}}void torpu(){for (int i = 1; i <= cnt; i++){if (indegree[i] == 0){q.push(i);}}while (!q.empty()){int cur = q.front();q.pop();if (col[cur] == 0){col[cur] = 1;col[cf[cur]] = -1;}for (int i = headtwo[cur]; i != -1; i = edgetwo[i].next){int v = edgetwo[i].v;if (--indegree[v] == 0){q.push(v);}}}}void solve(){for (int i = 0; i < 2 * n; i++){if (!dfn[i]){tarjan(i);}}bool flag = true;for (int i = 0; i < n; i++){if (belong[i] == belong[i + n]){flag = false;}cf[belong[i]] = belong[i + n];cf[belong[i + n]] = belong[i];}if (!flag){cout << "bad luck" << endl;}else{for (int i = 0; i < e; i++){if (belong[edge[i].u] != belong[edge[i].v]){insert(belong[edge[i].v], belong[edge[i].u]);indegree[belong[edge[i].u]]++;}}torpu();int num = 0;for (int i = 1; i < n; i++){if (col[belong[i]] == 1){cout << i << 'h';}else{cout << i << 'w';}if (i != n - 1){cout << ' ';}}cout << endl;}}void input(){int u, v;char ch1, ch2;while (cin >> n >> m, n + m){init();for (int i = 0; i < m; i++){cin >> u >> ch1 >> v >> ch2;if (v == 0 && ch2 == 'w'){continue;}if (ch1 == 'w' && ch2 == 'w'){add(u, v + n);add(v, u + n);}else if (ch1 == 'w' && ch2 == 'h'){add(u, v);add(v + n, u + n);}else if (ch1 == 'h' && ch2 == 'w'){add(v, u);add(u + n, v + n);}else{add(u + n, v);add(v + n, u);}}add(0, 0 + n);solve();}}int main(){input();return 0;}

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