Poj3678：Katu Puzzle

最新推荐文章于 2021-11-07 12:46:17 发布

原创最新推荐文章于 2021-11-07 12:46:17 发布 · 287 阅读

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本文介绍了一种基于2-SAT算法解决布尔表达式满足性问题的方法。通过构建图模型并运用Tarjan算法进行强连通分量分析，该文详细阐述了如何判断给定条件下是否存在可行解，并附带完整代码实现。

大概题意

有 $n$ 个数，可以为 $0/1$ ，给 $m$ 个条件，表示某两个数经过 $or, and, xor$ 后的数是多少
判断是否有解

Sol

$2-SAT$ 判定
建图

# include <iostream>
# include <stdio.h>
# include <stdlib.h>
# include <string.h>
# include <math.h>
# include <algorithm>
# define RG register
# define IL inline
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long ll;
const int _(2005);
const int __(4e6 + 5);

IL int Input(){
    RG int x = 0, z = 1; RG char c = getchar();
    for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
    for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
    return x * z;
}

int n, m, first[_], cnt, num;
int S[_], vis[_], dfn[_], low[_], Index, col[_];
struct Edge{
    int to, next;
} edge[__];

IL void Add(RG int u, RG int v){
    edge[cnt] = (Edge){v, first[u]}; first[u] = cnt++;
}

IL void Tarjan(RG int u){
    vis[u] = 1, dfn[u] = low[u] = ++Index, S[++S[0]] = u;
    for(RG int e = first[u]; e != -1; e = edge[e].next){
        RG int v = edge[e].to;
        if(!dfn[v]) Tarjan(v), low[u] = min(low[u], low[v]);
        else if(vis[v]) low[u] = min(low[u], dfn[v]);
    }
    if(dfn[u] != low[u]) return;
    RG int v = S[S[0]--]; col[v] = ++num, vis[v] = 0;
    while(v != u) v = S[S[0]--], col[v] = num, vis[v] = 0;
}

int main(RG int argc, RG char* argv[]){
    Fill(first, -1), n = Input(), m = Input();
    for(RG int i = 1; i <= m; ++i){
        RG int u = Input() + 1, v = Input() + 1, w = Input();
        RG char op; scanf(" %c", &op);
        if(op == 'A'){
            if(w) Add(u, v), Add(v, u), Add(u + n, u), Add(v + n, v);
            else Add(u, v + n), Add(v, u + n);
        }
        else if(op == 'O'){
            if(w) Add(u + n, v), Add(v + n, u);
            else Add(u, u + n), Add(v, v + n), Add(u + n, v + n), Add(v + n, u + n);
        }
        else{
            if(w) Add(u, v + n), Add(v, u + n), Add(u + n, v), Add(v + n, u);
            else Add(u, v), Add(v, u), Add(u + n, v + n), Add(v + n, u + n);
        }
    }
    for(RG int i = 1, tmp = n << 1; i <= tmp; ++i)
        if(!dfn[i]) Tarjan(i);
    for(RG int i = 1; i <= n; ++i)
        if(col[i] == col[i + n]) return puts("NO"), 0;
    return puts("YES"), 0;
}