TTTTTTTTTTTTTT POJ 3678 与或异或 2-SAT+强连通 模板题

Katu Puzzle

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 9129		Accepted: 3391

Description

Katu Puzzle is presented as a directed graph G(V, E) with each edge e(a, b) labeled by a boolean operator op (one of AND, OR, XOR) and an integer c (0 ≤ c ≤ 1). One Katu is solvable if one can find each vertex V_i a value X_i (0 ≤ X_i ≤ 1) such that for each edge e(a, b) labeled by op and c, the following formula holds:

 X_a op X_b = c

The calculating rules are:

AND	0	1
0	0	0
1	0	1

OR	0	1
0	0	1
1	1	1

XOR	0	1
0	0	1
1	1	0

Given a Katu Puzzle, your task is to determine whether it is solvable.

Input

The first line contains two integers N (1 ≤ N ≤ 1000) and M,(0 ≤ M ≤ 1,000,000) indicating the number of vertices and edges.
The following M lines contain three integers a (0 ≤ a < N), b(0 ≤ b < N), c and an operator op each, describing the edges.

Output

Output a line containing "YES" or "NO".

Sample Input

4 4
0 1 1 AND
1 2 1 OR
3 2 0 AND
3 0 0 XOR

Sample Output

YES

Hint

X₀ = 1, X₁ = 1, X₂ = 0, X₃ = 1.

有一个有向图G（V，E），每条边e（a,b）上有一个位运算符op（AND， OR或XOR）和一个值c（0或1）。

问能不能在这个图上的每个点分配一个值X（0或1），使得每一条边e(a,b)满足  Xa op Xb =  c

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <algorithm>
#include <set>
using namespace std;
typedef long long ll;
typedef unsigned long long Ull;
#define MM(a,b) memset(a,b,sizeof(a));
const double eps = 1e-10;
const int  inf =0x7f7f7f7f;
const double pi=acos(-1);
const int maxn=10+1000;

int n,m,pre[maxn],lowlink[maxn],dfs_clock,sccno[maxn],scc_cnt;
vector<int> G[2*maxn];
stack<int> S;

void read()
{
    int a,b,c;char s[10];
    scanf("%d %d %d %s",&a,&b,&c,s);
    if(s[0]=='A')
    {
        if(c)
        {
           G[a].push_back(a+n);
           G[b].push_back(b+n);
        }
        else
        {
           G[a+n].push_back(b);
           G[b+n].push_back(a);
        }
    }
    else if(s[0]=='O')
    {
        if(!c)
        {
            G[a+n].push_back(a);
            G[b+n].push_back(b);
        }
        else
        {
            G[a].push_back(b+n);
            G[b].push_back(a+n);
        }
    }
    else if(s[0]=='X')
    {
        if(c)
        {
            G[a].push_back(b+n);
            G[a+n].push_back(b);
            G[b].push_back(a+n);
            G[b+n].push_back(a);
        }
        else
        {
            G[a].push_back(b);
            G[a+n].push_back(b+n);
            G[b].push_back(a);
            G[b+n].push_back(a+n);
        }
    }
}

void tarjan(int u)
{
    pre[u]=lowlink[u]=++dfs_clock;
    S.push(u);
    for(int i=0;i<G[u].size();i++)
    {
        int v=G[u][i];
        if(!pre[v])
            {
                tarjan(v);
                lowlink[u]=min(lowlink[u],lowlink[v]);
            }
        else if(!sccno[v])
                lowlink[u]=min(lowlink[u],pre[v]);
    }

    if(lowlink[u]==pre[u])
    {
        scc_cnt++;
        while(1)
        {
            int x=S.top();S.pop();
            sccno[x]=scc_cnt;
            if(x==u) break;//找到了当前强连通的起始节点就退出，<br>//不然会破坏其他强连通分量
        }
    }
}

void find_scc()
{
    MM(pre,0);
    MM(sccno,0);
    scc_cnt=dfs_clock=0;
    for(int i=0;i<n;i++)
      if(!pre[i])
            tarjan(i);
}

int main()
{
    while(~scanf("%d %d",&n,&m))
    {
        for(int i=0;i<2*n;i++) G[i].clear();
        for(int i=1;i<=m;i++)
              read();

        find_scc();

        int flag=1;
        for(int i=0;i<n;i++)
            if(sccno[i]==sccno[i+n])
          {
            flag=0;
            break;
          }
        if(flag) printf("YES
");
        else printf("NO
");
    }
    return 0;
}

　　2-sat的建图参考；

http://blog.csdn.net/u011466175/article/details/23048459?utm_source=tuicool&utm_medium=referral

和http://blog.csdn.net/leolin_/article/details/7215871