POJ 3648 Wedding

2-SAT问题，麻烦就麻烦再需要输出解具体见赵爽论文。

可以判断有无解时，用一个映射，把与自己相对的“强联通分量”用hash映射出来，即hash[belong[i]] = belong[i+n]等等。这样在拓扑序染色时就比较方便了。

#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <string>
#include <cstring>
#include <cmath>
#include <vector>
#include <queue>
#include <algorithm>
#include <map>
using namespace std;

const int maxn = 1010;
const int maxm = 1010*1010;

struct Edge
{
    int v, w;
    int next;
}edge[maxm], edge2[maxm];

int first2[maxn];
int cnt2;
int first[maxn], stack[maxn], ins[maxn], dfn[maxn], low[maxn];
int belong[maxn];

int n, m, k;
int cnt;
int scnt, top, tot;

void init()
{
    cnt = 0;
    cnt2 = 0;
    scnt = top = tot = 0;
    memset(first, -1, sizeof(first));
    memset(first2, -1, sizeof(first2));
    memset(dfn, 0, sizeof(dfn));
    memset(ins, 0, sizeof(ins));
    memset(low, 0, sizeof(low));
}

void read_graph(int u, int v)
{
    edge[cnt].v = v;
    edge[cnt].next = first[u], first[u] = cnt++;
}

void read_graph2(int u, int v)
{
    edge2[cnt2].v = v;
    edge2[cnt2].next = first2[u], first2[u] = cnt2++;
}

void dfs(int u)
{
    int v;
    low[u] = dfn[u] = ++tot;
    stack[top++] = u;
    ins[u] = 1;
    for(int e = first[u]; e != -1; e = edge[e].next)
    {
        v = edge[e].v;
        if(!dfn[v]) { dfs(v); low[u] = min(low[u], low[v]); }
        else if(ins[v]) { low[u] = min(low[u], dfn[v]); }
    }
    if(low[u] == dfn[u])
    {
        scnt++;
        do { v = stack[--top]; belong[v] = scnt; ins[v] = 0; } while(u != v);
    }
}

void readint(int &x)
{
    char c;
    while(!isdigit(c)) c = getchar();
    
    x = 0;
    while(isdigit(c))
    {
        x = x*10 + c-'0';
        c = getchar();
    }
}

void writeint(int x)
{
    if(x > 9) writeint(x/10);
    putchar(x%10+'0');
}

void Tarjan()
{
    for(int v = 0; v < 2*n; v++) if(!dfn[v])
        dfs(v);
}

void read_case()
{
    init();
    
    int x, y;
    char s1[10], s2[10];
    while(m--)
    {
        int a, b;
        scanf("%d%s %d%s", &x, s1, &y, s2);
        if(s1[0] == 'h' && s2[0] == 'h')
        {
            read_graph(2*x+1, 2*y);
            read_graph(2*y+1, 2*x);
        }
        if(s1[0] == 'h' && s2[0] == 'w')
        {
            read_graph(2*x+1, 2*y+1);
            read_graph(2*y, 2*x);
        }
        if(s1[0] == 'w' && s2[0] == 'h')
        {
            read_graph(2*x, 2*y);
            read_graph(2*y+1, 2*x+1);
        }
        if(s1[0] == 'w' && s2[0] == 'w')
        {
            read_graph(2*x, 2*y+1);
            read_graph(2*y, 2*x+1);
        }
    }
    read_graph(0, 1);
}

int hash[maxn];

int check()
{
    for(int i = 0; i < 2*n; i+=2)
    {
        if(belong[i] == belong[i+1]) return 0;
        else hash[belong[i]] = belong[i+1], hash[belong[i+1]] = belong[i];
    }
    return 1;
}

int color[maxn];
int ind[maxn];

void toposort()
{
    queue<int> Q;
    
    for(int i = 1; i <= scnt; i++) if(!ind[i])
        Q.push(i);
    while(!Q.empty())
    {
        int u = Q.front(); Q.pop();
        if(!color[u]) color[u] = 1, color[hash[u]] = 2;
        for(int e = first2[u]; e != -1; e = edge2[e].next)
        {
            int v = edge2[e].v;
            if(--ind[v] == 0) Q.push(v);
        }
    }
}

void build()
{
    memset(ind, 0, sizeof(ind));
    memset(color, 0, sizeof(color));
    
    for(int u = 0; u < 2*n; u++)
    {
        for(int e = first[u]; e != -1; e = edge[e].next)
        {
            int v = edge[e].v;
            if(belong[u] != belong[v])
            {
                read_graph2(belong[v], belong[u]);
                ind[belong[u]]++;
            }
        }
    }
    
    toposort();
}

void output()
{
    for(int i = 2; i < 2*n; i++)
    {
        if(color[belong[i]] == color[belong[0]])
        {
            if(i & 1) printf("%dh ", i/2);
            else printf("%dw ", i/2);
        }
    }
    printf("
");
}

void solve()
{
    read_case();
    
    Tarjan();
    
    if(check())
    {
        build();
        output();
    }
    else printf("bad luck
");
}

int main()
{
    while(~scanf("%d%d", &n, &m) && (n || m))
    {
        solve();
    }
    return 0;
}