「Bzoj 1823」「JSOI2010」满汉全席 (2-SAT)

BZOJ 1823
经典2-SAT模型，注意输入

Tarjan版本：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<stack>
#define ms(i, j) memset(i, j, sizeof i)
#define LL long long
#define db double
using namespace std;
int tb, n, m, belongs[2 * 100 + 5], vis[2 * 100 + 5], scc_num, dfn[2 * 100 + 5], low[2 * 100 + 5];
vector<int> G[2 * 100 + 5];//2x->m 2x+1->h
stack<int> s;
void tarjan(int u) {
    dfn[u] = low[u] = ++tb, vis[u] = -1, s.push(u);
    for (int i = 0; i < (int)G[u].size(); i++) {
        int v = G[u][i];
        if (vis[v] == 0) {
            tarjan(v);
            low[u] = min(low[u], low[v]);
        } else if (vis[v] == -1) low[u] = min(low[u], dfn[v]);
    }
    if (low[u] == dfn[u]) {
        int e;
        scc_num++;
        do {
            e = s.top(); s.pop();
            belongs[e] = scc_num;
            vis[e] = 1;
        } while (e != u);
    }
}
int getss()
{
    char ans;
    while (true) {
        ans = getchar();
        if (ans == 'm') return 0;
        if (ans == 'h') return 1;
        int ret = 0;
        while (ans >= '0' && ans <= '9')  {
            ret = ret * 10 + ans - '0';
            ans = getchar();
        }
        if (ret > 0) return ret;
    }
}
void clean() {
    tb = scc_num = 0;
    for (int i = 0; i <= 2 * 100 + 1; i++) dfn[i] = low[i] = belongs[i] = vis[i] = 0, G[i].clear();
}
void solve() {
    scanf("%d%d", &n, &m);
    clean();
    char c1, c2;
    int a1, a2;
    for (int i = 1; i <= m; i++) {
        c1 = getss(), a1 = getss(), c2 = getss(), a2 = getss();
        int x = 2 * (a1 - 1) + (c1 == 1) , y = 2 * (a2 - 1) + (c2 == 1);
        G[x].push_back(y ^ 1), G[y].push_back(x ^ 1);
    }
    for (int i = 1; i <= n * 2; i++) if (!vis[i]) tarjan(i);
    for (int i = 1; i <= 2 * n; i += 2) {
        if (belongs[i] == belongs[i ^ 1]) {
            printf("BAD\n");
            return ;
        }
    }
    printf("GOOD\n");
} 
int main() {
    int T; scanf("%d", &T);
    while (T--) solve();
    return 0;
}

#include<cstdio>    
#include<algorithm>    
#include<cstring>  
#include<vector>   
#define ms(i,j) memset(i,j, sizeof i);    
using namespace std;
const int MAXN = 100 + 5;
struct twoSAT
{
    bool mark[MAXN*2];
    vector<int> G[MAXN*2];
    int n;
    int c, S[MAXN*2];
    void init(int ni)
    {
        n = ni;
        for (int i=0;i<2*n;i++) G[i].clear();
        ms(mark, false);
    }
    bool dfs(int x)
    {
        if (mark[x^1]) return false;
        if (mark[x]) return true;
        mark[x] = true;
        S[++c] = x;
        for (int i=0;i<G[x].size();i++)
        if (!dfs(G[x][i])) return false;
        return true;
    }
    void addE(int x, int y, int xv, int yv)
    {
        x = x * 2 + xv;
        y = y * 2 + yv;
        G[x^1].push_back(y);
        G[y^1].push_back(x);//2x for man, 2x+1 for han
    }
    bool solve()
    {
        for (int i=0;i<2*n;i+=2)
        if (!mark[i] && !mark[i+1])
        {
            c = 0;
            if (!dfs(i))
            {
                for (int j=1;j<=c;j++) mark[S[j]] = false;
                c = 0;
                if (!dfs(i+1)) return false;
            }
        }
        return true;
    }
}ts;
int k;
int n,m;
int getss()
{
    char ans;
    while (true)
    {
        ans = getchar();
        if (ans=='m') return 0;
        if (ans=='h') return 1;
        int ret = 0;
        while (ans>='0'&&ans<='9') 
        {
            ret = ret*10+ans-'0';
            ans = getchar();
        }
        if (ret>0) return ret;
    }
}
void inpu()
{
    int v1 = getss();
    int z1 = getss();
    int v2 = getss();
    int z2 = getss();
    ts.addE(z1-1, z2-1, v1, v2);
}
int main()   
{     
    scanf("%d", &k);
    while (k--)
    {
        scanf("%d%d", &n,&m);
        ts.init(n);
        for (int i=0;i<m;i++) 
        {
            inpu();
        }
        if (ts.solve())
        {
            printf("GOOD\n");
        } else printf("BAD\n");
    }
    return 0;    
}