【UOJ #210】【UER #6】寻找罪犯

题目描述

有n个人分为好人和坏人，说了m句话。好人不会说假话，坏人至多说一句谎话。求出一组解，满足要求。

题解

利用2-SAT拆点，一个人拆成两个点，表示他是好人和坏人。然而这样的话边数是m^2的，所以用前/后缀和优化构图即可。

#include <bits/stdc++.h>
using namespace std;

const int N = 1e6 + 10;
const int M = 3e6 + 10;
inline int read()
{
    char ch = getchar();int x = 0, f = 1;
    while(ch < '0' || ch > '9') {if(ch == '-') f = -1; ch = getchar();}
    while(ch >= '0' && ch <= '9') {x = (x << 1) + (x << 3) - '0' + ch; ch = getchar();}
    return x * f;
}
int n, m;
int h[N], e[M], ne[M], idx;
int dfn[N], low[N], num, in[N], stac[N], top, pre[N];
int c[N], cnt;

void add(int a, int b)
{
    e[idx] = b;
    ne[idx] = h[a];
    h[a] = idx++;
}

int crm(int x, int f)
{
    return x + n * f;
}

int word(int x, int f)
{
    return n + n + x + m * f;
}

void tarjan(int u)
{
    stac[++ top] = u;
    dfn[u] = low[u] = ++ num;
    in[u] = 1;
    for (int i = h[u]; ~i; i = ne[i])
    {
        int j = e[i];
        if(!dfn[j])
        {
            tarjan(j);
            low[u] = min(low[u], low[j]);
        }
        else if(in[j])
            low[u] = min(low[u], dfn[j]);
    }
    if(low[u] == dfn[u])
    {
        int z;
        cnt ++;
        do{
            z = stac[top --];
            c[z] = cnt;
            in[z] = 0;
        }while(z != u);
    }
}


int main()
{
    memset(h, -1, sizeof(h));
    n = read(), m = read();
    for (int i = 1; i <= n; i ++) pre[i] = 2*m + 1;
    for (int i = 1; i <= m; i ++)
    {
        int x = read(), y = read(), t = read();
        t ^= 1;
        add(crm(y, t ^ 1), word(pre[x], 0)); //若这次y的身份与x说的相反，则x前面说的话都对。
        
        add(word(pre[x], 1), crm(y, t));    //若x前面说过谎话了，y的身份与x说的相符

        add(word(i, 0), crm(y, t));         //若说的话一直是真的，y的身份与x相符
        
        add(crm(y, t ^ 1), word(i, 1));     //若y的身份与x说的不符，x这句话是假的
        
        add(word(pre[x], 1), word(i, 1));   //前面说过假话，现在也保持说过假话的状态
        
        add(word(i, 0), word(pre[x], 0));   //现在都没说过假话，前面也没说过假话
        pre[x] = i;                         //更新前缀
    }
    for (int i = 1; i <= n; i ++)
    {
        add(word(pre[i], 1), crm(i, 1));    //说过假话，一定是坏人
        add(crm(i, 0), word(pre[i], 0));    //好人一定没说过假话
    }
    for (int i = 1; i <= (n + m) * 2; i ++)
            if(!dfn[i])
                tarjan(i);
    bool flag = true;
    for (int i = 1; i <= n; i ++)
    {
        if(c[i] == c[i + n])
        {
            flag = false;
            break;
        }
    }
    for (int i = 1; i <= m; i ++)
    {
        if(c[n + n + i] == c[n + n + m + i])
        {
            flag = false;
            break;
        }
    }
    if(!flag)
    {
        puts("Impossible");
    }
    else
    {
        vector<int>ans;
        for (int i = 1; i <= n; i ++)
        {
            if(c[i] > c[i + n])
                ans.push_back(i);
        }
        cout << ans.size() << '\n';
        for (int i = 0; i < ans.size(); i ++)
        {
            printf("%d%c", ans[i], i == ans.size() -1?'\n' : ' ');
        }
    }
}