BZOJ 3495 [2-SAT]

$Description$

$k$个国家，$n$个城市，$m$条边。要求每个国家有且仅有一个首都，每条边两端的城市至少要有一个首都。判断是否有解, 有解输出“TAK”，无解输出”NIE”。$1≤k, n ,m≤1000000$。

$Solution$

经典二元关系 2-SAT 的模型，但是关系数是 O(n 2 ) 的，不
能承受。
对于一个有 y 个城市的国家，新建 2y 个变量，分别表示前
i 个城市和后 i 个城市中是否有首都。
如此一来关系数降至 O(n + m)。
时间复杂度 O(n + m)。
by Claris

#include <cstdio>
#include <iostream>
using namespace std;

const int N = 6010101;

inline char get(void) {
    static char buf[100000], *S = buf, *T = buf;
    if (S == T) {
        T = (S = buf) + fread(buf, 1, 100000, stdin);
        if (S == T) return EOF;
    }
    return *S++;
}
inline void read(int &x) {
    static char c; x = 0; int sgn = 0;
    for (c = get(); c < '0' || c > '9'; c = get()) if (c == '-') sgn = 1;
    for (; c >= '0' && c <= '9'; c = get()) x = x * 10 + c - '0';
    if (sgn) x = -x;
}


int k, n, m, Gcnt, x, y, p, Scnt, cnt, dlc, top;
int c[N];
int S[N];
struct edge {
    int to, next;
    edge(int t = 0, int n = 0):to(t), next(n) {}
};
edge G[N * 2];
int head[N], pre[N], low[N], scc[N];

inline void AddEdge(int from, int to) {
    G[++Gcnt] = edge(to, head[from]); head[from] = Gcnt;
}
inline int Min(int a, int b) {
    return a < b ? a : b;
}
void dfs(int u) {
    pre[u] = low[u] = ++dlc;
    int to; S[++top] = u;
    for (int i = head[u]; i; i = G[i].next) {
        to = G[i].to;
        if (!pre[to]) {
            dfs(to); low[u] = Min(low[u], low[to]);
        } else if (!scc[to]) {
            low[u] = Min(low[u], pre[to]);
        }
    }
    if (low[u] == pre[u]) {
        Scnt++; int x;
        while (true) {
            x = S[top]; top--;
            scc[x] = Scnt;
            if (x == u) break;
        }
    }
}
bool Check(void) {
    int lim = cnt << 1 | 1;
    for (int i = 1; i <= lim; i++)
        if (!pre[i]) dfs(i);
    for (int i = 1; i <= cnt; i++)
        if (scc[i << 1] == scc[i << 1 | 1]) return false;
    return true;
}

int main(void) {
    read(n); read(m); read(k);
    for (int i = 0; i < m; i++) {
        read(x); read(y);
        AddEdge(x << 1, y << 1 | 1);
        AddEdge(y << 1, x << 1 | 1);
    }
    cnt = n;
    while (k--) {
        read(p); 
        for (int i = 1; i <= p; i++) {
            read(x); y = cnt + i;
            if (i != 1) {
                AddEdge((y - 1) << 1 | 1, y << 1 | 1);
                AddEdge(y << 1, (y - 1) << 1);
                AddEdge(x << 1 | 1, (y - 1) << 1);
            } else {
                AddEdge(x << 1, y << 1);
            }
            AddEdge(y << 1, x << 1);
            AddEdge(x << 1 | 1, y << 1 | 1);
            y = cnt + i + p;
            if (i != p) {
                AddEdge((y + 1) << 1 | 1, y << 1 | 1);
                AddEdge(y << 1, (y + 1) << 1);
                AddEdge(x << 1 | 1, (y + 1) << 1);
            } else {
                AddEdge(x << 1, y << 1);
            }
            AddEdge(y << 1, x << 1);
            AddEdge(x << 1 | 1, y << 1 | 1);
        }
        cnt += p * 2;
    }
    if (Check()) printf("TAK\n");
    else printf("NIE\n");
    return 0;
}