pku3678 Katu Puzzle 2-sat判断是否存在可行解

http://poj.org/problem?id=3678

题意：

给定n个点，这些点只能取0或1。然后给出m条边，每条边四个变量 a,b,c,op  op的取值为（AND,OR，XOR） 问是否存在一组解X0，X1,....Xn-1使得每条边满足Xa op Xb = c  Xa,Xb表示每条边的端点。

思路：

2-sat。 将每个顶点i拆分成两个点，2*i和2*i +1 分表表示0，1。然后根据已知条件建图。建图的思想不是很清晰给出过程：

(1) A and B = 0 添加弧 A->!B , B->!A
(2)A and B = 1 !A->A , !B->B
(3)A or B = 0 A->!A , B->!B
(4)A or B = 1 !A->B , !B->A
(5)A xor B = 0 A->B , B->A , !A->!B ,!B->!A
(6)A xor B = 1 A->!B, B->!A,!B->A,!A->B

然后就是2-sat判断有无解了。

//#pragma comment(linker,"/STACK:327680000,327680000")
#include <iostream>
#include <cstdio>
#include <cmath>
#include <vector>
#include <cstring>
#include <algorithm>
#include <string>
#include <set>
#include <functional>
#include <numeric>
#include <sstream>
#include <stack>
#include <map>
#include <queue>

#define CL(arr, val)    memset(arr, val, sizeof(arr))

#define ll long long
#define inf 0x7f7f7f7f
#define lc l,m,rt<<1
#define rc m + 1,r,rt<<1|1
#define pi acos(-1.0)
#define ll long long
#define L(x)    (x) << 1
#define R(x)    (x) << 1 | 1
#define MID(l, r)   (l + r) >> 1
#define Min(x, y)   (x) < (y) ? (x) : (y)
#define Max(x, y)   (x) < (y) ? (y) : (x)
#define E(x)        (1 << (x))
#define iabs(x)     (x) < 0 ? -(x) : (x)
#define OUT(x)  printf("%I64d\n", x)
#define lowbit(x)   (x)&(-x)
#define Read()  freopen("din.txt", "r", stdin)
#define Write() freopen("dout.txt", "w", stdout);


#define N 2007
#define M 1000007
using namespace std;

struct node
{
    int v;
    int next;
}g[N*N];
int head[N],ct;

int dfn[N],low[N];
int belong[N],stk[N];
bool isn[N];
int idx,cnt,top;

int n,m;
char op[10];

void add(int u,int v)
{
    g[ct].v = v;
    g[ct].next = head[u];
    head[u] = ct++;
}
//建图是关键
void build(int u,int v,int c,char *op)
{
    if (op[0] == 'A')
    {
        if (c == 1)
        {
            add(2*u,2*u + 1);
            add(2*v,2*v + 1);
        }
        else
        {
            add(2*u + 1,2*v);
            add(2*v + 1,2*u);
        }
    }
    else if (op[0] == 'O')
    {
        if (c == 1)
        {
            add(2*u,2*v + 1);
            add(2*v,2*u + 1);
        }
        else
        {
            add(2*u + 1,2*u);
            add(2*v + 1,2*v);
        }
    }
    else if (op[0] == 'X')
    {
        if (c == 1)
        {
            add(2*u + 1,2*v);
            add(2*v,2*u + 1);
            add(2*v + 1,2*u);
            add(2*u,2*v + 1);
        }
        else
        {
            add(2*u + 1,2*v + 1);
            add(2*v + 1,2*u + 1);
            add(2*u,2*v);
            add(2*v,2*u);
        }
    }
}
void init()
{
    int i;
    for (i = 0; i < 2*n; ++i)
    {
        dfn[i] = low[i] = -1;
        belong[i] = 0;
        isn[i] = false;
    }
    idx = cnt = top = 0;
}
void tarjan(int u)
{
    int i,j;
    dfn[u] = low[u] = ++idx;
    stk[++top] = u;
    isn[u] = true;
    for (i = head[u]; i != -1; i = g[i].next)
    {
        int v = g[i].v;
        if (dfn[v] == -1)
        {
            tarjan(v);
            low[u] = min(low[u],low[v]);
        }
        else if (isn[v])
        {
            low[u] = min(low[u],dfn[v]);
        }
    }
    if (dfn[u] == low[u])
    {
        cnt++;
        do
        {
            j = stk[top--];
            isn[j] = false;
            belong[j] = cnt;
        } while (j != u);
    }
}
void solve()
{
    int i;
    init();
    for (i = 0; i < 2*n; ++i)
    {
        if (dfn[i] == -1) tarjan(i);
    }
    bool flag = false;
    for (i = 0; i < n; ++i)
    {
        if (belong[2*i] == belong[2*i + 1])
        {
            flag = true;
            break;
        }
    }
    if (flag) printf("NO\n");
    else printf("YES\n");
}
int main()
{
    //Read();
    int i ;
    int x,y,z;
    while (~scanf("%d%d",&n,&m))
    {
        CL(head,-1); ct = 0;
        for (i = 0; i < m; ++i)
        {
            scanf("%d%d%d%s",&x,&y,&z,op);
            build(x,y,z,op);
        }
        solve();
    }
    return 0;
}