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  • poj 3678 Katu Puzzle(Two Sat)

    题目链接:http://poj.org/problem?id=3678

    代码:

    #include<cstdio>
    #include<cstring>
    #include<iostream>
    #include<algorithm>
    #include<vector>
    using namespace std;
    
    const int maxn = 1050;
    
    struct Two_Sat
    {
        int n;
        vector<int> G[maxn*2];
        bool mark[maxn*2];
        int s[maxn*2],cnt;
    
        void init(int n)
        {
            this->n = n;
            memset(mark,0,sizeof(mark));
            for(int i=0; i<n*2; i++) G[i].clear();
        }
    
        void add_clause(int u,int v,int flag,char symbol)
        {
            u = u*2 + flag;
            v = v*2 + flag;
            if(symbol == 'A')
            {
                if(flag)
                {
                    G[u].push_back(v);
                    G[v].push_back(u);
                    G[u^1].push_back(u);
                    G[v^1].push_back(v);
                }
                else
                {
                    G[u^1].push_back(v);
                    G[v^1].push_back(u);
                }
            }
            else if(symbol == 'O')
            {
                if(flag)
                {
                    G[u^1].push_back(v);
                    G[v^1].push_back(u);
                }
                else
                {
                    G[u].push_back(v);
                    G[v].push_back(u);
                    G[u^1].push_back(u);
                    G[v^1].push_back(v);
                }
            }
            else
            {
                if(flag)
                {
                    G[u^1].push_back(v);
                    G[v].push_back(u^1);
                    G[v^1].push_back(u);
                    G[u].push_back(v^1);
                }
                else
                {
                    G[u].push_back(v);
                    G[v].push_back(u);
                    G[u^1].push_back(v^1);
                    G[v^1].push_back(u^1);
                }
            }
        }
    
        bool dfs(int u)
        {
            if(mark[u^1])   return false;
            if(mark[u])     return true;
            mark[u] = true;
            s[cnt++] = u;
            for(int i=0; i<G[u].size(); i++)
            {
                if(!dfs(G[u][i])) return false;
            }
            return true;
        }
    
        bool solve()
        {
            for(int i=0; i<n*2; i+=2)
            {
                if(!mark[i] && !mark[i+1])
                {
                    cnt = 0;
                    if(!dfs(i))
                    {
                        while(cnt>0) mark[s[--cnt]] = false;
                        if(!dfs(i+1))   return false;
                    }
                }
            }
            return true;
        }
    }solver;
    
    
    int main()
    {
        //freopen("E:\acm\input.txt","r",stdin);
        int N,M;
        while(cin>>N>>M)
        {
            solver.init(N);
            for(int i=1; i<=M; i++)
            {
                int a,b,c;
                char s[5];
                scanf("%d %d %d %s",&a,&b,&c,s);
                solver.add_clause(a,b,c,s[0]);
            }
            if(solver.solve())  printf("YES
    ");
            else                printf("NO
    ");
        }
    }
    View Code
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  • 原文地址:https://www.cnblogs.com/acmdeweilai/p/3348308.html
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