【2-sat】P4171 [JSOI2010] 满汉全席

P4171 [JSOI2010] 满汉全席 - 洛谷

大意:n个点 m个条件 形如m1,h32，满足n个条件

思路:2-sat 让m=0，h=1 ，然后转换为i m j h的建图，注意傻逼题目的数字可能是多位数 不能直接x[1]-'0'

#include <cstdio>
#include <stack>
#include <iostream>
#include <cstring>
#include <cmath>
#include <algorithm>
#define ep emplace_back 

#define ios std::ios::sync_with_stdio(false);std::cin.tie(0); 
const int N = 1e5+9;

using namespace std;

int trans(char x){
    return x == 'm' ? 0 : 1;
}
//转换函数 把字符串的数字转换成数字
int num(string x){
    int ans=0;
    int len  = x.size()-1;
    for(int i=1 ; i<x.size();++i){
        int ch =x[i]-'0';
        ans+=ch*pow(10,len-1);
        --len;
    }
    return ans;
}
int n, m;
int head[N], idx = 0;
int time__;
int low[N], dfn[N];
struct node{
    int to, val, next;
} e[N<<2];
bool vis[N];
int scc[N], scc_cnt = 0;

void add(int u, int v, int val){
    e[idx] = {v, val, head[u]};
    head[u] = idx++;
}

void bd(){
    cin >> n >> m;
    for(int k = 1; k <= m; ++k){
        string ch1, ch2;
        cin >> ch1 >> ch2;
        int i = num(ch1);
        int j = num(ch2);
        int a = trans(ch1[0]);
        int b = trans(ch2[0]);
        //printf("%d %d\n",i,j);
        add(i + a * n, j + !b * n, 0);       
        add(j + b * n, i + !a * n, 0);
    }
}

stack<int> st;
void tarjan(int u){
    low[u] = dfn[u] = ++time__;
    st.push(u);
    vis[u] = 1;
    for(int i = head[u]; i != -1; i = e[i].next){
        int v = e[i].to;
        if(!dfn[v]){
            tarjan(v);
            low[u] = min(low[u], low[v]);
        }
        else if(vis[v]){
            low[u] = min(low[u], dfn[v]);
        }
    }
    if(dfn[u] == low[u]){
        scc_cnt++;
        while(true){
            int v = st.top();
            st.pop();
            vis[v] = 0;
            scc[v] = scc_cnt;
            if(u == v) break;
        }
    }
}

void clear(){
    memset(head, -1, sizeof head);
    memset(scc, 0, sizeof scc);
    memset(vis, 0, sizeof vis);
    memset(low, 0, sizeof low);
    memset(dfn, 0, sizeof dfn);
    while(!st.empty()) st.pop();
    time__ = 0, idx = 0, scc_cnt = 0;
}

void solve(){
    clear();
    bd();
    
    for(int i = 1; i <= 2 * n; ++i)
        if(!dfn[i])
            tarjan(i);
    bool ok = 1;
    for(int i = 1; i <= n; ++i){
        if(scc[i] == scc[i + n]){
            ok = 0;
            break;
        }
    }
   /* */
    cout << (ok ? "GOOD" : "BAD") ;
    cout<< "\n";
}

int main(){
    ios;
    int T = 1;
    cin >> T;
    while(T--){
        solve();
    }
    return 0;
}