[JSOI 2010] 满汉全席

最新推荐文章于 2025-12-18 11:54:58 发布

转载最新推荐文章于 2025-12-18 11:54:58 发布 · 185 阅读

0 ·

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/evenbao/p/9508025.html

文章标签：

#php

本文深入解析了2-SAT算法的原理及其在解决特定逻辑问题中的应用。通过一个具体的编程实例，详细展示了如何使用2-SAT算法来判断一组命题逻辑公式是否具有一致的解决方案。代码中包含了深度优先搜索（DFS）和Tarjan算法实现强连通分量（SCC）的计算，以此来确定是否存在冲突的命题组合。

[题目链接]

https://www.lydsy.com/JudgeOnline/problem.php?id=1823

[算法]

2-SAT

[代码]

#include<bits/stdc++.h>
using namespace std;
#define MAXN 110
#define MAXM 1010

struct edge
{
        int to,nxt;
} e[MAXM << 1];

int timer,cnt,top,tot;
int dfn[MAXN << 1],low[MAXN << 1],belong[MAXN << 1],s[MAXN << 1],head[MAXN << 1];
bool instack[MAXN];

template <typename T> inline void read(T &x)
{
        int f = 1; x = 0;
        char c = getchar();
        for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
        for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + c - '0';
        x *= f;
}
inline void addedge(int u,int v)
{
        tot++;
        e[tot] = (edge){v,head[u]};
        head[u] = tot; 
}
inline void tarjan(int u)
{
        low[u] = dfn[u] = ++timer;
        instack[u] = true;
        s[++top] = u;
        for (int i = head[u]; i; i = e[i].nxt)
        {
                int v = e[i].to;
                if (!dfn[v])
                {
                        tarjan(v);
                        low[u] = min(low[u],low[v]);
                } else if (instack[v]) low[u] = min(low[u],dfn[v]);
        }
        if (low[u] == dfn[u])
        {
                cnt++;
                while (s[top + 1] != u)
                {
                        instack[s[top]] = false;
                        belong[s[top]] = cnt;
                        top--;
                }
        }
}
int main() 
{
        
        int T;
        read(T);
        while (T--)        
        {
                int n,m;
                read(n); read(m);
                tot = 0;
                for (int i = 1; i <= 2 * n; i++) head[i] = 0;
                for (int i = 1; i <= m; i++)
                {
                        char a[20] , b[20];
                        scanf("%s%s",a + 1,b + 1);
                        int t1 = 0 , t2 = 0;
                        for (int j = 2; j <= strlen(a + 1); j++) t1 = t1 * 10 + a[j] - '0';
                        for (int j = 2; j <= strlen(b + 1); j++) t2 = t2 * 10 + b[j] - '0';
                        if (a[1] == 'm') addedge(t1 + n,t2 + (b[1] == 'h') * n);
                        else addedge(t1,t2 + (b[1] == 'h') * n);
                        if (b[1] == 'm') addedge(t2 + n,t1 + (a[1] == 'h') * n);
                        else addedge(t2,t1 + (a[1] == 'h') * n);
                } 
                timer = cnt = 0;
                for (int i = 1; i <= 2 * n; i++) low[i] = dfn[i] = 0;
                for (int i = 1; i <= 2 * n; i++)
                {
                        if (!dfn[i])
                                tarjan(i);
                }
                bool ans = true;
                for (int i = 1; i <= n; i++)
                {
                      if (belong[i] == belong[i + n]) 
                      {
                              ans = false;
                              break;
                        }
                }
                if (ans) printf("GOOD\n");
                else printf("BAD\n");
        }
        
        return 0;
    
}