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Language: Default Katu Puzzle
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 Xa op Xb = c The calculating rules are:
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. Source |
题意：n个点给出m个关系（AND,OR,XOR），问是否存在解。
思路：２-SAT建图
a AND b = 1: ~x->x,~y->y （两个数必须全为1）
a AND b = 0: y->~x,x->~y （两个数至少有一个为0）
a OR b = 1：~x->y,~y->x （两个数至少有一个为1）
a OR b = 0: x->~x,y->~y （两个数必须全为0）
a XOR b = 1:x->~y,y->~x,~y->x,~x->y （两个数必须不同）
a XOR b = 0:x->y,y->x,~x->~y,~y->~x （两个数必须相同）
代码：
#include <iostream>
#include <functional>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#pragma comment (linker,"/STACK:102400000,102400000")
#define pi acos(-1.0)
#define eps 1e-6
#define lson rt<<1,l,mid
#define rson rt<<1|1,mid+1,r
#define FRE(i,a,b) for(i = a; i <= b; i++)
#define FREE(i,a,b) for(i = a; i >= b; i--)
#define FRL(i,a,b) for(i = a; i < b; i++)
#define FRLL(i,a,b) for(i = a; i > b; i--)
#define mem(t, v) memset ((t) , v, sizeof(t))
#define sf(n) scanf("%d", &n)
#define sff(a,b) scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define pf printf
#define DBG pf("Hi\n")
typedef long long ll;
using namespace std;
#define INF 0x3f3f3f3f
#define mod 1000000009
const int maxn = 1005;
const int MAXN = 3005;
const int MAXM = 1200010;
const int N = 1005;
struct Edge
{
int to,next;
}edge[MAXM];
int tot,head[MAXN];
int Low[MAXN],DFN[MAXN],Stack[MAXN],Belong[MAXN];
bool Instack[MAXN];
int top,Index,scc;
int n,m;
char str[10];
void init()
{
tot=0;
memset(head,-1,sizeof(head));
}
void addedge(int u,int v)
{
edge[tot].to=v;
edge[tot].next=head[u];
head[u]=tot++;
}
void Tarjan(int u)
{
int v;
Low[u]=DFN[u]=++Index;
Stack[top++]=u;
Instack[u]=true;
for (int i=head[u];~i;i=edge[i].next)
{
v=edge[i].to;
if (!DFN[v])
{
Tarjan(v);
if (Low[u]>Low[v])
Low[u]=Low[v];
}
else if (Instack[v]&&Low[u]>DFN[v])
Low[u]=DFN[v];
}
if (Low[u]==DFN[u])
{
++scc;
do{
v=Stack[--top];
Instack[v]=false;
Belong[v]=scc;
}while (v!=u);
}
return ;
}
bool solvable(int n)
{
memset(DFN,0,sizeof(DFN));
memset(Instack,false,sizeof(Instack));
Index=scc=top=0;
for (int i=0;i<n;i++)
if (!DFN[i])
Tarjan(i);
for (int i=0;i<n;i+=2)
{
if (Belong[i]==Belong[i^1])
return false;
}
return true;
}
int main()
{
#ifndef ONLINE_JUDGE
freopen("C:/Users/lyf/Desktop/IN.txt","r",stdin);
#endif
int i,j,a,b,c;
scanf("%d%d",&n,&m);
init();
for (i=0;i<m;i++)
{
scanf("%d%d%d %s",&a,&b,&c,str);
if (strcmp(str,"OR")==0)
{
if (c==0)
{
addedge(2*a,2*a+1);
addedge(2*b,2*b+1);
}
else
{
addedge(2*a+1,2*b);
addedge(2*b+1,2*a);
}
}
else if (strcmp(str,"AND")==0)
{
if (c==0)
{
addedge(2*a,2*b+1);
addedge(2*b,2*a+1);
}
else
{
addedge(2*a+1,2*a);
addedge(2*b+1,2*b);
}
}
else
{
if (c==0)
{
addedge(2*a,2*b);
addedge(2*a+1,2*b+1);
addedge(2*b,2*a);
addedge(2*b+1,2*a+1);
}
else
{
addedge(2*a,2*b+1);
addedge(2*a+1,2*b);
addedge(2*b,2*a+1);
addedge(2*b+1,2*a);
}
}
}
if (solvable(2*n))
printf("YES\n");
else
printf("NO\n");
return 0;
}
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