Wedding (poj 3648 2-SAT 输出随意一组解)

Posted 2020-09-30 lxjshuju

Language: Default Wedding Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 9004   Accepted: 2722   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 Source Waterloo Local Contest, 2007.9.29

题意：题目太绕了=-=有一对新人结婚。非常多对夫妇參加婚礼，共n对。如今安排全部人坐在一张桌子两边，且（1）每对夫妇不能坐在同一側（2）n对夫妇中有通奸关系（包含男男，男女，女女），有通奸关系的不能坐在新娘的对面。问是否存在可行的安排方案。若存在输出与新娘同側的人。

思路：注意新郎必须在还有一边，所以就要加一条新娘到新郎的边以满足情况。

代码：

#include <iostream>
#include <functional>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#pragma comment (linker,"/STACK:102400000,102400000")
#define pi acos(-1.0)
#define eps 1e-6
#define lson rt<<1,l,mid
#define rson rt<<1|1,mid+1,r
#define FRE(i,a,b)  for(i = a; i <= b; i++)
#define FREE(i,a,b) for(i = a; i >= b; i--)
#define FRL(i,a,b)  for(i = a; i < b; i++)
#define FRLL(i,a,b) for(i = a; i > b; i--)
#define mem(t, v)   memset ((t) , v, sizeof(t))
#define sf(n)       scanf("%d", &n)
#define sff(a,b)    scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define pf          printf
#define DBG         pf("Hi\n")
typedef long long ll;
using namespace std;

#define INF 0x3f3f3f3f
#define mod 1000000009
const int maxn = 1005;
const int MAXN = 100;
const int MAXM = 2010;
const int N = 1005;

struct Edge
{
    int to,next;
}edge[MAXM];

int n,m;
int tot,head[MAXN];
int Low[MAXN],DFN[MAXN],Stack[MAXN],Belong[MAXN];
bool Instack[MAXN];
int top,Index,scc;

void init()
{
    tot=0;
    memset(head,-1,sizeof(head));
}

void addedge(int u,int v)
{
    edge[tot].to=v;
    edge[tot].next=head[u];
    head[u]=tot++;
}

void Tarjan(int u)
{
    int v;
    Low[u]=DFN[u]=++Index;
    Instack[u]=true;
    Stack[top++]=u;
    for (int i=head[u];~i;i=edge[i].next)
    {
        v=edge[i].to;
        if (!DFN[v])
        {
            Tarjan(v);
            if (Low[u]>Low[v]) Low[u]=Low[v];
        }
        else if (Instack[v]&&Low[u]>DFN[v])
            Low[u]=DFN[v];
    }
    if (Low[u]==DFN[u])
    {
        scc++;
        do
        {
            v=Stack[--top];
            Instack[v]=false;
            Belong[v]=scc;
        }while (v!=u);
    }
    return ;
}

bool solvable(int n)
{
    memset(DFN,0,sizeof(DFN));
    memset(Instack,false,sizeof(Instack));
    top=Index=scc=0;
    for (int i=0;i<n;i++)
        if (!DFN[i])
            Tarjan(i);
    for (int i=0;i<n;i+=2)
    {
        if (Belong[i]==Belong[i^1])
            return false;
    }
    return true;
}

queue<int>q1;
vector<vector<int> >dag;
char color[MAXN];
int indeg[MAXN];
int cf[MAXN];

void solve(int n)
{
    dag.assign(scc+1,vector<int>());
    memset(indeg,0,sizeof(indeg));
    memset(color,0,sizeof(color));
    for (int u=0;u<n;u++)
    {
        for (int i=head[u];~i;i=edge[i].next)
        {
            int v=edge[i].to;
            if (Belong[u]!=Belong[v])
            {
                dag[Belong[v]].push_back(Belong[u]);
                indeg[Belong[u]]++;
            }
        }
    }
    for (int i=0;i<n;i++)
    {
        cf[Belong[i]]=Belong[i^1];
        cf[Belong[i^1]]=Belong[i];
    }
    while (!q1.empty()) q1.pop();
    for (int i=1;i<=scc;i++)
        if (indeg[i]==0)
            q1.push(i);
    while (!q1.empty())
    {
        int u=q1.front();
        q1.pop();
        if (color[u]==0)
        {
            color[u]='R';
            color[cf[u]]='B';
        }
        int sz=dag[u].size();
        for (int i=0;i<sz;i++)
        {
            indeg[dag[u][i]]--;
            if (indeg[dag[u][i]]==0)
                q1.push(dag[u][i]);
        }
    }
}

int main()
{
#ifndef ONLINE_JUDGE
    freopen("C:/Users/lyf/Desktop/IN.txt","r",stdin);
#endif
    int i,j,a,b;
    char c1,c2;
    while (~scanf("%d%d",&n,&m))
    {
        if (n==0&&m==0) break;
        init();
        for (i=0;i<m;i++)
        {
            scanf("%d%c%d%c",&a,&c1,&b,&c2);
            if (c1=='w') a=2*a; else a=2*a+1;
            if (c2=='w') b=2*b; else b=2*b+1;
            addedge(a,b^1);
            addedge(b,a^1);
        }
        addedge(0,1);
        if (!solvable(2*n))
        {
            printf("bad luck\n");
            continue;
        }
        solve(2*n);
        for (i=1;i<n;i++)
        {
            if (i>1) printf(" ");
            if (color[Belong[2*i]]=='R') printf("%dh",i);
            else printf("%dw",i);
        }
        printf("\n");
    }
    return 0;
}