[JSOI2010]满汉全席 2-SAT

https://www.luogu.org/problemnew/show/P4171

意识到图中只有两种不同的菜系：满和汉

并且检查员类似于一个约束，可以发现这就是一个2-sat模型，满和汉分别对应true和false

由于只是检查可行性，只需要判断存在点的true个false存在同一个强连通分量即可。

#include <map>
#include <set>
#include <ctime>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <string>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;
#define For(i, x, y) for(int i=x;i<=y;i++)
#define _For(i, x, y) for(int i=x;i>=y;i--)
#define Mem(f, x) memset(f,x,sizeof(f))
#define Sca(x) scanf("%d", &x)
#define Sca2(x,y) scanf("%d%d",&x,&y)
#define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define Scl(x) scanf("%lld",&x);
#define Pri(x) printf("%d\n", x)
#define Prl(x) printf("%lld\n",x);
#define CLR(u) for(int i=0;i<=N;i++)u[i].clear();
#define LL long long
#define ULL unsigned long long
#define mp make_pair
#define PII pair<int,int>
#define PIL pair<int,long long>
#define PLL pair<long long,long long>
#define pb push_back
#define fi first
#define se second
typedef vector<int> VI;
int read(){int x = 0,f = 1;char c = getchar();while (c<‘0‘ || c>‘9‘){if (c == ‘-‘) f = -1;c = getchar();}
while (c >= ‘0‘&&c <= ‘9‘){x = x * 10 + c - ‘0‘;c = getchar();}return x*f;}
const double eps = 1e-9;
const int maxn = 210;
const int maxm = 2010;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7;
int N,M,K;
struct Edge{
    int to,next;
}edge[maxm << 2];
int head[maxn],tot;
int Low[maxn],dfn[maxn],Stack[maxn],Belong[maxn];
int Index,top,scc;
bool Instack[maxn];
void init(){
    for(int i = 0 ; i <= (N << 1) ; i ++){
        head[i] = -1;
        Low[i] = dfn[i] = Belong[i] = Instack[i] = 0;
    }
    tot = scc = top = Index = 0;
}
void add(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;
    Stack[top++] = u;
    Instack[u] = true;
    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);
    }
}
int main(){
//    ios::sync_with_stdio(false);
    int T = read();
    while(T--){
        cin >> N >> M; init();
        for(int i = 1; i <= M ; i ++){
            char op1,op2; int x,y;
            cin >> op1 >> x >> op2 >> y;
            int flag1 = op1 == ‘m‘?1:0;
            int flag2 = op2 == ‘m‘?1:0;
            add(x + N * (flag1 ^ 1),y + N * (flag2 & 1));
            add(y + N * (flag2 ^ 1),x + N * (flag1 & 1));
        }
        for(int i = 1; i <= (N << 1) ; i ++) if(!dfn[i]) Tarjan(i);
        int flag = 1;
        for(int i = 1; i <= N && flag; i ++){
            if(Belong[i] == Belong[i + N]) flag = 0;
        }
        if(flag) puts("GOOD");
        else puts("BAD");
    }
    return 0;
}

原文地址：https://www.cnblogs.com/Hugh-Locke/p/10774101.html