Katu Puzzle
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 6702 | Accepted: 2463 |
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 Vi a value Xi (0 ≤ Xi ≤ 1) such that for each edge e(a, b) labeled by op and c, the following formula holds:
Xa op Xb = c
The calculating rules are:
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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.
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真正理解了2-SAT才能建出图
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/** head-file **/
#include <iostream>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <list>
#include <set>
#include <map>
#include <algorithm>
/** define-for **/
#define REP(i, n) for (int i=0;i<int(n);++i)
#define FOR(i, a, b) for (int i=int(a);i<int(b);++i)
#define DWN(i, b, a) for (int i=int(b-1);i>=int(a);--i)
#define REP_1(i, n) for (int i=1;i<=int(n);++i)
#define FOR_1(i, a, b) for (int i=int(a);i<=int(b);++i)
#define DWN_1(i, b, a) for (int i=int(b);i>=int(a);--i)
#define REP_N(i, n) for (i=0;i<int(n);++i)
#define FOR_N(i, a, b) for (i=int(a);i<int(b);++i)
#define DWN_N(i, b, a) for (i=int(b-1);i>=int(a);--i)
#define REP_1_N(i, n) for (i=1;i<=int(n);++i)
#define FOR_1_N(i, a, b) for (i=int(a);i<=int(b);++i)
#define DWN_1_N(i, b, a) for (i=int(b);i>=int(a);--i)
/** define-useful **/
#define clr(x,a) memset(x,a,sizeof(x))
#define sz(x) int(x.size())
#define see(x) cerr<<#x<<" "<<x<<endl
#define se(x) cerr<<" "<<x
#define pb push_back
#define mp make_pair
/** test **/
#define Display(A, n, m) {
REP(i, n){
REP(j, m) cout << A[i][j] << " ";
cout << endl;
}
}
#define Display_1(A, n, m) {
REP_1(i, n){
REP_1(j, m) cout << A[i][j] << " ";
cout << endl;
}
}
using namespace std;
/** typedef **/
typedef long long LL;
/** Add - On **/
const int direct4[4][2]={ {0,1},{1,0},{0,-1},{-1,0} };
const int direct8[8][2]={ {0,1},{1,0},{0,-1},{-1,0},{1,1},{1,-1},{-1,1},{-1,-1} };
const int direct3[6][3]={ {1,0,0},{0,1,0},{0,0,1},{-1,0,0},{0,-1,0},{0,0,-1} };
const int MOD = 1000000007;
const int INF = 0x3f3f3f3f;
const long long INFF = 1LL << 60;
const double EPS = 1e-9;
const double OO = 1e15;
const double PI = acos(-1.0); //M_PI;
const int maxn=11111;
const int maxm=2111111;
int n,m;
struct EDGENODE{
int to;
int next;
};
struct TWO_SAT{
int head[maxn*2];
EDGENODE edges[maxm*2];
int edge;
int n;
void init(int n){
this->n=2*n;
clr(head,-1);
edge=0;
}
void addedge(int u,int v){
edges[edge].to=v,edges[edge].next=head[u],head[u]=edge++;
}
// x = xval or y = yval
//!x->y,!y->x
void add_clause(int x,int xval,int y,int yval){
x=x*2+xval;
y=y*2+yval;
addedge(x^1,y);
addedge(y^1,x);
}
//x=xval
//!x=x
void add_con(int x,int xval){
x=x*2+xval;
addedge(x^1,x);
}
//--
void add_self(int x,int xval,int y,int yval){
x=x*2+xval;
y=y*2+yval;
addedge(x,y);
}
int pre[maxn],lowlink[maxn],sccno[maxn],scc_cnt,dfs_clock;
stack<int>stk;
void dfs(int u)
{
pre[u]=lowlink[u]=++dfs_clock;
stk.push(u);
for (int i=head[u];i!=-1;i=edges[i].next){
int v=edges[i].to;
if (!pre[v]){
dfs(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++;
int x;
do{
x=stk.top();
stk.pop();
sccno[x]=scc_cnt;
}while (x!=u);
}
}
void find_scc(int n)
{
dfs_clock=scc_cnt=0;
clr(sccno,0);
clr(pre,0);
while (!stk.empty()) stk.pop();
REP(i,n) if (!pre[i]) dfs(i);
}
bool solve(){
find_scc(n);
for (int i=0;i<n;i+=2){
if (sccno[i]==sccno[i^1]) return false;
}
return true;
}
}TwoSAT;
int main()
{
int a,b,c;
char s[4];
while (~scanf("%d%d",&n,&m))
{
TwoSAT.init(n);
REP(i,m)
{
scanf("%d%d%d%s",&a,&b,&c,s);
/*
if (!strcmp(s,"AND")){
if (c==1){
TwoSAT.add_self(a,1,a,0);
TwoSAT.add_self(b,1,b,0);
}else if (c==0){
TwoSAT.add_self(a,0,b,1);
TwoSAT.add_self(b,0,a,1);
}
}
if (!strcmp(s,"OR")){
if (c==1){
TwoSAT.add_self(a,1,b,0);
TwoSAT.add_self(b,1,a,0);
}else if (c==0){
TwoSAT.add_self(a,0,a,1);
TwoSAT.add_self(b,0,b,1);
}
}
if (!strcmp(s,"XOR")){
if (c==1){
TwoSAT.add_self(a,0,b,1);
TwoSAT.add_self(b,0,a,1);
TwoSAT.add_self(a,1,b,0);
TwoSAT.add_self(b,1,a,0);
}else if (c==0){
TwoSAT.add_self(a,0,b,0);
TwoSAT.add_self(b,0,a,0);
TwoSAT.add_self(a,1,b,1);
TwoSAT.add_self(b,1,a,1);
}
}*/
if (!strcmp(s,"AND")){
if (c==1){
TwoSAT.add_con(a,1);
TwoSAT.add_con(b,1);
}else if (c==0){
TwoSAT.add_clause(a,0,b,0);
}
}
if (!strcmp(s,"OR")){
if (c==1){
TwoSAT.add_clause(a,1,b,1);
}else if (c==0){
TwoSAT.add_con(a,0);
TwoSAT.add_con(b,0);
}
}
if (!strcmp(s,"XOR")){
if (c==1){
TwoSAT.add_clause(a,1,b,1);
TwoSAT.add_clause(a,0,b,0);
}else if (c==0){
TwoSAT.add_clause(a,1,b,0);
TwoSAT.add_clause(a,0,b,1);
}
}
}
if (TwoSAT.solve()) puts("YES");
else puts("NO");
}
return 0;
}