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;
}