K-TV Show Game(2-SAT)

传送门

题意

题目要求n个灯(R,B),给出m组赋值方式,每一组有三个赋值，至少有两个是正确的,问是否能找到一组正确的赋值方式.（摘自此博客）

分析

如何转化为2_SAT?
设3 B 4 R 5 R：
如果3为R，则4比为R，5比为R
如果3为B，则3比为B，5比为R
如果5为B，则3比为B，4比为R

如此转化，每一组赋值可以建立6个约束关系（单向边）
最后套板子输出可行解就行了

代码

#include <bits/stdc++.h>

using namespace std;
//-----pre_def----
const double PI = acos(-1.0);
const int INF = 0x3f3f3f3f;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> PII;
typedef pair<double, double> PDD;
#define fir(i, a, b) for (int i = (a); i <= (b); i++)
#define rif(i, a, b) for (int i = (a); i >= (b); i--)
#define endl '\n'
#define init_h memset(h, -1, sizeof h), idx = 0;
#define lowbit(x) x &(-x)
//---------------
const int N = 1e4 + 10;
int low[N], tim[N], tot, vis[N], top, id[N], sk[N], pick[N], n, dp[N];
vector<int> G[N];
int cnt;
void tarjan(int x)
{
    tim[x] = low[x] = ++tot;
    sk[++top] = x;
    vis[x] = 1;
    for (int i = 0; i < G[x].size(); i++)
    {
        int v = G[x][i];
        if (tim[v] == 0)
        {
            tarjan(v);
            low[x] = min(low[x], low[v]);
        }
        else if (vis[v])
            low[x] = min(low[x], low[v]);
    }
    if (low[x] == tim[x])
    {
        int v;
        ++cnt;
        while (1)
        {
            v = sk[top--];
            vis[v] = 0;
            id[v] = cnt;
            if (v == x)
                break;
        }
    }
}

int main()
{
#ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
    freopen("out.txt", "w", stdout);
#endif
    int m;
    scanf("%d %d", &n, &m);
    for (int i = 1; i <= m; i++)
    {
        pair<int, char> d[4];
        fir(j, 1, 3)
        {
            scanf("%d %c", &d[j].first, &d[j].second);
            getchar();
        }
        //建6条边
        fir(j, 1, 3)
        {
            int v1 = d[j].first, v2 = d[j].first + n;
            //'B'==0,           'R'==1
            if (d[j].second == 'R')
                swap(v1, v2);
            for (int k = j + 1; k <= 3; k++)
            {
            //d[j] d[k]
                int w1 = d[k].first, w2 = d[k].first + n;
                if (d[k].second == 'B')
                    swap(w1, w2);
                if (w1 != v1)
                    G[w1].push_back(v1); 
                if (v2 != w2)
                    G[v2].push_back(w2);
            }
        }
    }
    for (int i = 1; i <= 2 * n; i++)
        if (tim[i] == 0)
            tarjan(i);

    for (int i = 1; i <= n; i++)
        if (id[i] == id[i + n])
        {
            printf("-1\n");
            return 0;
        }

    for (int i = 1; i <= n; i++)
    {
        if (id[i] < id[i + n])
            printf("B");
        else
            printf("R");
    }
    printf("\n");
    return 0;
}