poj 3648 Wedding

類型：2-sat

題目：http://poj.org/problem?id=3648

來源：Waterloo Local Contest, 2007.9.29

思路：每個人都有可能坐在左邊或者右邊，將其分爲兩個點i， i + 2 * n。定義新娘在左側。

（1）對於一對有曖昧關係的人i, j。如果i在右邊，必有j在左邊，如果j在右邊，必有i在左邊。構造有向邊： addedge1(i + 2 * n, j);  addedge1(j + 2 * n, i)。

（2）夫婦需一左一右

然後通過2-sat算法判斷可行性，如果可行並輸出解。

！！！新娘必在左邊，新郎必在右邊。構造有向邊 addedge1(2 * n + 1, 1); addedge1(n + 1, 3 * n + 1);

// poj 3648 Wedding
// wa wa ac 196K 0MS
#include <iostream>
#include <string>
#include <queue>
#include <stack>
#include <algorithm>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
using namespace std;

#define FOR(i,a,b) for(i = (a); i < (b); ++i)
#define FORE(i,a,b) for(i = (a); i <= (b); ++i)
#define FORD(i,a,b) for(i = (a); i > (b); --i)
#define FORDE(i,a,b) for(i = (a); i >= (b); --i)
#define CLR(a,b) memset(a,b,sizeof(a))
#define PB(x) push_back(x)

const int N = 200;
const int M = 40010;

int low[N], dfn[N];
int belong[N];
bool inStack[N], vis[N];
stack<int> st;
int n, m;
int step, t;
int conflict[N];//縮點後有矛盾的點
int du[N];//入度
int color[N];//1爲紅色，-1爲藍色，紅色輸出
int cnt1, cnt2;
int head1[N], head2[N];
int top[N], cnt;
struct node {
    int s, e;
    int len;
}c[N];
struct edge1 {
    int v, nxt;
}E1[M];
struct edge2 {
    int v, nxt;
}E2[M];

void addedge1(int u, int v) {
    E1[cnt1].v = v;
    E1[cnt1].nxt = head1[u];
    head1[u] = cnt1++;
}

void addedge2(int u, int v) {
    E2[cnt2].v = v;
    E2[cnt2].nxt = head2[u];
    head2[u] = cnt2++;
}

void tarjan(int u) {
    int i;
    step++;
    st.push(u);
    low[u] = dfn[u] = step;
    vis[u] = 1;
    inStack[u] = 1;
    for(i = head1[u]; i != -1; i = E1[i].nxt) {
        int x = E1[i].v;
        if(!vis[x]) {
            tarjan(x);
            low[u] = min(low[u], low[x]);
        }
        else if(inStack[x])
            low[u]=min(low[u], dfn[x]);
    }
    if(low[u] == dfn[u]) {
        t++;
        while(1) {
            int x = st.top();
            st.pop();
            belong[x] = t;
            inStack[x] = 0;
            if(x == u)
                break;
        }
    }
}

void init() {
    cnt1 = cnt2 = step = t = 0;
    CLR(head1, -1);
    CLR(head2, -1);
    CLR(du, 0);
    CLR(color, 0);
    CLR(vis, 0);
    CLR(inStack, 0);
}

bool isConflict(int a, int b, int c, int d) {
    if(a >= d || b <= c)
        return false;
    return true;
}

void Rebuild(){//逆圖
    int i, j;

    FORE(i, 1, 4 * n) {
        for(j = head1[i]; j != -1; j = E1[j].nxt) {
            int a = belong[i], b = belong[E1[j].v];
            if(a != b) {
                addedge2(b, a);
                du[a]++;
            }
        }
    }
    FORE(i, 1, 2 * n) {
        int a = belong[i], b = belong[i + 2 * n];
        conflict[a] = b;
        conflict[b] = a;
    }
}

void topsort() {
    int i,j;

    queue<int> q;
    FORE(i, 1, t)
        if(du[i] == 0)
            q.push(i);
    cnt = 0;
    while(!q.empty()) {
        int x = q.front();
        top[++cnt] = x;
        q.pop();
        for(i = head2[x]; i != -1; i = E2[i].nxt) {
            int tmp = E2[i].v;
            du[tmp]--;
            if(du[tmp] == 0)
                q.push(tmp);
        }
    }
}

void dfs_Blue(int u) {
    int i;
    color[u] = -1;
    for(i = head2[u]; i != -1; i = E2[i].nxt) {
        int x = E2[i].v;
        if(color[x] == 0)
            dfs_Blue(x);
    }
}

void dfs_Red() {
    int i, j;

    FORE(i, 1, cnt) {
        int x = top[i];
        if(color[x] == 0) {
            color[x] = 1;
            dfs_Blue(conflict[x]);
        }
    }
}

void output() {
    int i, j, t1, t2;

    int len = 0;
    int sign = color[belong[1]];
    FORE(i, 2, 2 * n) {
        int x = belong[i], y = belong[i + 2 * n];
        if(color[x] == sign) {
            if(i <= n)
                (len++ == 0) ? printf("%dw", i - 1) : printf(" %dw", i - 1);
            else
                (len++ == 0) ? printf("%dh", i - n - 1) : printf(" %dh", i - n - 1);
        }
    }
    printf("\n");
}

void solve() {
    int i, j;
    FORE(i, 1, 4 * n)
        if(!vis[i])
            tarjan(i);
    FORE(i, 1, 2 * n) {
        if(belong[i] == belong[i + 2 * n]) {
            printf("bad luck\n");
            return ;
        }
    }
    Rebuild();
    topsort();
    dfs_Red();
    output();
}

int main() {
    char str1[5], str2[5];

    while(scanf("%d %d", &n, &m) != EOF, n || m) {
        init();
        int i, j, t1, t2;
        FORE(i, 1, m) {
            scanf("%d%c%d%c", &t1, &str1[0], &t2, &str2[0]);
            ++t1, ++t2;
            if(str1[0] == 'h')
                t1 += n;
            if(str2[0] == 'h')
                t2 += n;
            addedge1(t1 + 2 * n, t2);
            addedge1(t2 + 2 * n, t1);
        }
        addedge1(2 * n + 1, 1);
        addedge1(n + 1, 3 * n + 1);
        FORE(i, 1, n) {
            addedge1(i, i + 3 * n);
            addedge1(i + 2 * n, i + n);
            addedge1(i + n, i + 2 * n);
            addedge1(i + 3 * n, i);
        }
        solve();
    }
    return 0;
}