HDU - 3062 - Party(2-SAT)

链接：

https://vjudge.net/problem/HDU-3062

题意：

有n对夫妻被邀请参加一个聚会，因为场地的问题，每对夫妻中只有1人可以列席。在2n 个人中，某些人之间有着很大的矛盾（当然夫妻之间是没有矛盾的），有矛盾的2个人是不会同时出现在聚会上的。有没有可能会有n 个人同时列席？

思路：

2-SAT模板，考虑两个不能同时存在的夫妻，他们的对象可以同时存在，加到边上。
训练指南的模板
tarjan模板，
把所有必需的边连上，缩点，如果某个夫妻再同一个强联通分量内，就不能满足

代码：

//#include<bits/stdc++.h>
#include<iostream>
#include<string>
#include<cstdio>
#include<vector>
#include<string.h>
#include<cstring>
#include<set>
#include<queue>
#include<algorithm>
#include<math.h>
#include<stdio.h>
#include<map>
#include<stack>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
const int MOD = 20071027;
const int MAXN = 1e3+10;
int Next[4][2] = {-1, 0, 0, 1, 1, 0, 0, -1};

struct TwoSAT
{
    int n;
    vector<int> G[MAXN*2];
    bool mark[MAXN*2];
    int S[MAXN*2], c;

    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 < (int)G[x].size();i++)
            if (!dfs(G[x][i])) return false;
        return true;
    }

    void init(int n)
    {
        this->n = n;
        for (int i = 0;i < 2*n;i++)
            G[i].clear();
        memset(mark, 0, sizeof(mark));
    }

    void add_clause(int x, int xval, int y, int yval)
    {
        // x = xval or y = yval
        x = x*2+xval;
        y = y*2+yval;
        G[x^1].push_back(y);
        G[y^1].push_back(x);
    }

    bool solve()
    {
        for (int i = 0;i < n*2;i += 2)
        {
            if (!mark[i] && !mark[i^1])
            {
                c = 0;
                if (!dfs(i))
                {
                    while(c > 0) mark[S[--c]] = false;
                    if (!dfs(i^1)) return false;
                }
            }
        }
        return true;
    }
}sat;

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);
    int n, m;
    while(cin >> n >> m)
    {
        sat.init(n);
        int x1, x2, y1, y2;
        for (int i = 1;i <= m;i++)
        {
            cin >> x1 >> x2 >> y1 >> y2;
            sat.add_clause(x1, y1^1, x2, y2^1);
        }
        if (sat.solve())
            cout << "YES" << endl;
        else
            cout << "NO" << endl;
    }

    return 0;
}

tarjan算法模板

//tarjan
//#include<bits/stdc++.h>
#include<iostream>
#include<string>
#include<cstdio>
#include<vector>
#include<string.h>
#include<cstring>
#include<set>
#include<queue>
#include<algorithm>
#include<math.h>
#include<stdio.h>
#include<map>
#include<stack>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
const int MOD = 20071027;
const int MAXN = 1e3+10;
int Next[4][2] = {-1, 0, 0, 1, 1, 0, 0, -1};

vector<int> G[MAXN*2];
stack<int> St;
int dfn[MAXN*2], low[MAXN*2], sccnum[MAXN*2];
int dfn_clock, scc_cnt;
int n, m;

void tarjan(int u)
{
    dfn[u] = low[u] = ++dfn_clock;
    St.push(u);
    for (int i = 0;i < (int)G[u].size();i++)
    {
        int v = G[u][i];
        if (!dfn[v])
        {
            tarjan(v);
            low[u] = min(low[v], low[u]);
        }
        else if (!sccnum[v])
            low[u] = min(dfn[v], low[u]);
    }
    if (low[u] == dfn[u])
    {
        ++scc_cnt;
        while(true)
        {
            int x = St.top();
            St.pop();
            sccnum[x] = scc_cnt;
            if (x == u)
                break;
        }
    }
}

bool solve()
{
    memset(dfn, 0, sizeof(dfn));
    memset(low, 0, sizeof(low));
    memset(sccnum, 0, sizeof(sccnum));
    dfn_clock = scc_cnt = 0;
    for (int i = 0;i < 2*n;i++)
        if (!dfn[i]) tarjan(i);
    for (int i = 0;i < 2*n;i+=2)
        if (sccnum[i] == sccnum[i+1]) return false;
    return true;
}

void add_edge(int x, int xval, int y, int yval)
{
    x = x*2+xval;
    y = y*2+yval;
    G[x^1].push_back(y);
    G[y^1].push_back(x);
}

void init()
{
    for (int i = 0;i < 2*n;i++)
        G[i].clear();
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);
    while(cin >> n >> m)
    {
        init();
        int x1, x2, y1, y2;
        for (int i = 1;i <= m;i++)
        {
            cin >> x1 >> x2 >> y1 >> y2;
            add_edge(x1, y1^1, x2, y2^1);
        }
        if (solve())
            cout << "YES" << endl;
        else
            cout << "NO" << endl;
    }

    return 0;
}