Eliminate the Conflict HDU - 4115（2-sat 建图 hhh）

Posted 2021-01-13 wtsruvf

题意：

　　石头剪刀布 分别为1、2、3，有n轮，给出了小A这n轮出什么，然后m行，每行三个数a b k，如果k为0 表示小B必须在第a轮和第b轮的策略一样，如果k为1 表示小B在第a轮和第b轮的策略不一样，如果又一轮小B输了，那整个就输了，求小B能否战胜小A

解析：

　　写出来矛盾的情况  建图就好啦

可能我建的麻烦了。。。不过。。我喜欢 hhhhh

#include <iostream>
#include <cstdio>
#include <sstream>
#include <cstring>
#include <map>
#include <cctype>
#include <set>
#include <vector>
#include <stack>
#include <queue>
#include <algorithm>
#include <cmath>
#include <bitset>
#define rap(i, a, n) for(int i=a; i<=n; i++)
#define rep(i, a, n) for(int i=a; i<n; i++)
#define lap(i, a, n) for(int i=n; i>=a; i--)
#define lep(i, a, n) for(int i=n; i>a; i--)
#define rd(a) scanf("%d", &a)
#define rlld(a) scanf("%lld", &a)
#define rc(a) scanf("%c", &a)
#define rs(a) scanf("%s", a)
#define pd(a) printf("%d
", a);
#define plld(a) printf("%lld
", a);
#define pc(a) printf("%c
", a);
#define ps(a) printf("%s
", a);
#define MOD 2018
#define LL long long
#define ULL unsigned long long
#define Pair pair<int, int>
#define mem(a, b) memset(a, b, sizeof(a))
#define _  ios_base::sync_with_stdio(0),cin.tie(0)
//freopen("1.txt", "r", stdin);
using namespace std;
const int maxn = 1e5 + 10, INF = 0x7fffffff, LL_INF = 0x7fffffffffffffff;
int n, m;
int a[maxn];
vector<int> G[maxn];
int sccno[maxn], vis[maxn], low[maxn], scc_clock, scc_cnt;
stack<int> S;
void init()
{
    mem(sccno, 0);
    mem(vis, 0);
    mem(low, 0);
    scc_clock = scc_cnt = 0;
    for(int i = 0; i < maxn; i++) G[i].clear();
}

void dfs(int u)
{
    low[u] = vis[u] = ++scc_clock;
    S.push(u);
    for(int i = 0; i < G[u].size(); i++)
    {
        int v = G[u][i];
        if(!vis[v])
        {
            dfs(v);
            low[u] = min(low[u], low[v]);
        }
        else if(!sccno[v])
        {
            low[u] = min(low[u], vis[v]);
        }
    }
    if(low[u] == vis[u])
    {
        scc_cnt++;
        for(;;)
        {
            int x = S.top(); S.pop();
            sccno[x] = scc_cnt;
            if(x == u) break;
        }
    }
}

bool check()
{
    for(int i = 0; i < n * 2; i += 2)
        if(sccno[i] == sccno[i + 1])
        {
            return false;
        }
    return true;
}


int main()
{
    int T, kase = 0;
    cin >> T;
    while(T--)
    {
        init();
        int u, v, w;
        cin >> n >> m;
        for(int i = 0; i < n; i++)
            cin >> a[i];
        for(int i = 1; i <= m; i++)
        {
            cin >> u >> v >> w;
            u--, v--;
            int x = a[u], y = a[v];
            if(w == 1)
            {
                if(x == y)
                {
                    G[u << 1].push_back(v << 1 | 1);
                    G[v << 1 | 1].push_back(u << 1);
                    G[u << 1 | 1].push_back(v << 1);
                    G[v << 1].push_back(u << 1 | 1);
                }
                else if(x != y)
                {
                    if(x == 1 && y == 2 || y == 1 && x == 2)
                    {
                        if(y == 1 && x == 2)
                            swap(u, v);
                        G[u << 1 | 1].push_back(v << 1 | 1), G[v << 1].push_back(u << 1);
                    }
                    else if(x == 1 && y == 3 || y == 1 && x == 3)
                    {
                        if(y == 1 && x == 3)
                            swap(u, v);
                        G[v << 1 | 1].push_back(u << 1 | 1), G[u << 1].push_back(v << 1);
                    }
                    else if(x == 2 && y == 3 || x == 3 && y == 2)
                    {
                        if(x == 3 && y == 2)
                            swap(u, v);
                        G[u << 1 | 1].push_back(v << 1 | 1), G[v << 1].push_back(u << 1);
                    }
                }
            }
            else
            {
                if(x == y)
                {
                    G[u << 1].push_back(v << 1), G[u << 1 | 1].push_back(v << 1 | 1);
                    G[v << 1].push_back(u << 1), G[v << 1 | 1].push_back(u << 1 | 1);
                }
                else
                {
                    if(x == 1 && y == 2 || x == 2 && y == 1)
                    {
                        if(x == 2 && y == 1) swap(u, v);
                        G[u << 1 | 1].push_back(v << 1), G[v << 1].push_back(u << 1 | 1);
                        G[u << 1].push_back(u << 1 | 1), G[v << 1 | 1].push_back(v << 1);
                    }
                    else if(x == 1 && y == 3 || x == 3 && y == 1)
                    {
                        if(x == 3 && y == 1) swap(u, v);
                        G[u << 1].push_back(v << 1 | 1), G[v << 1 | 1].push_back(u << 1);
                        G[u << 1 | 1].push_back(u << 1), G[v << 1].push_back(v << 1 | 1);
                    }
                    else if(x == 2 && y == 3 || y == 2 && x == 3)
                    {
                        if(y == 2 && x == 3) swap(u, v);
                        G[u << 1 | 1].push_back(v << 1), G[v << 1].push_back(u << 1 | 1);
                        G[u << 1].push_back(u << 1 | 1), G[v << 1 | 1].push_back(v << 1);
                    }
                }
            }
        }
        for(int i = 0; i < n * 2; i++)
            if(!vis[i]) dfs(i);
        printf("Case #%d: ", ++kase);
        if(check())
        {
            cout << "yes" << endl;
        }
        else cout << "no" << endl;
    }

    return 0;
}