hdu 3678 Katu Puzzle (2-sat)

来源：互联网发布：苏州楼市成交数据编辑：程序博客网时间：2024/05/17 07:34

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:

AND01000101OR01001111XOR01001110

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 40 1 1 AND1 2 1 OR3 2 0 AND3 0 0 XOR

Sample Output

YES

Hint

X₀ = 1, X₁ = 1, X₂ = 0, X₃ = 1.

Source

POJ Founder Monthly Contest – 2008.07.27, Dagger

题意：有一个大小为N的集合={x1,x2..xn}，xi=0或1，现在给出它们之间的一些逻辑运算的结果(比如x1 and x2=1)，逻辑运算有AND OR XOR三种，问是否存在一种满足所有条件的取值方案。

思路：

每一位选择0或1两种状态，我们可以联想到2-sat；

对于建图还是挺简单的；

最近学会用两种2-sat算法；

第一种是TARJAN算法判强断联通分量求解，个人觉得局限性大；

第二种的话是搜索染色的方法，如果题目要输出方案，这种方法能够解决，并且输出字典序最小的结果；

下面用染色的方法做的；

代码：

#include<cstdio>#include<algorithm>#include<cstring>using namespace std;const int N=5005;const int M=4000005;struct node{    int now,next;}str[M];int head[N],tot,top;int vis[N],a[N];int select(char s[]){    if(s[0]=='A')        return 0;    else if(s[0]=='O')        return 1;    else        return 2;}void add(int x,int y){    str[tot].now=y;    str[tot].next=head[x];    head[x]=tot++;}bool dfs(int u,int n){    if(vis[u+n]) return false;    if(vis[u]) return true;    a[top++]=u;    vis[u]=1;    for(int i=head[u];i!=-1;i=str[i].next)    {        int v=str[i].now;        if(!dfs(v,n)) return false;    }    return true;}bool ok(int n){    memset(vis,0,sizeof(vis));    for(int i=0;i<n;i++)    {        if(!vis[i]&&!vis[i+n])        {            top=0;            if(!dfs(i,n))            {                while(top)                    vis[a[--top]]=0;                if(!dfs(i+n,n))                    return false;            }        }    }    return true;}int main(){    int n,m;    while(~scanf("%d%d",&n,&m))    {        tot=0;        memset(head,-1,sizeof(head));        for(int i=0;i<m;i++)        {            int a,b,c;            char s[5];            scanf("%d%d%d%s",&a,&b,&c,s);            if(select(s)==0)            {                if(c)                {                    add(a+n,b+n);                    add(b+n,a+n);                    add(a,a+n);                    add(b,b+n);                }                else                {                    add(a+n,b);                    add(b+n,a);                }            }            else if(select(s)==1)            {                if(c)                {                    add(a,b+n);                    add(b,a+n);                }                else                {                    add(a,b);                    add(b,a);                    add(a+n,a);                    add(b+n,b);                }            }            else            {                if(c)                {                    add(a,b+n);                    add(b,a+n);                    add(a+n,b);                    add(b+n,a);                }                else                {                    add(a,b);                    add(b,a);                    add(a+n,b+n);                    add(b+n,a+n);                }            }        }        if(ok(n))            printf("YES\n");        else            printf("NO\n");    }}

0 0