poj--3678--Katu Puzzle(2-sat 建模)

Katu Puzzle

Time Limit: 1000MS Memory Limit: 65536KB 64bit IO Format: %I64d & %I64u

SubmitStatus

Description

Katu Puzzle is presented as a directed graph G(V, E) with each edgee(a, b) labeled by a boolean operator op (one of AND, OR, XOR) and an integerc (0 ≤ c ≤ 1). One Katu is solvable if one can find each vertexV_i a value X_i (0 ≤ X_i ≤ 1) such that for each edgee(a, b) labeled by op and c, the following formula holds:

 X_aopX_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.

#include<stdio.h>#include<string.h>#include<queue>#include<stack>#include<vector>#include<algorithm>using namespace std;#define MAX 1000000+10int low[MAX],dfn[MAX];int sccno[MAX],m,n;int scc_cnt,dfs_clock;bool Instack[MAX];vector<int>G[MAX];stack<int>s;void init(){for(int i=0;i<2*n;i++)G[i].clear();}void getmap(){while(m--){int a,b,c;char op[5];memset(op,'\0',sizeof(op));scanf("%d%d%d%s",&a,&b,&c,op);if(op[0]=='A'){if(c==1){G[a+n].push_back(a);G[b+n].push_back(b);}else {G[a].push_back(b+n);G[b].push_back(a+n);}}else if(op[0]=='O'){if(c==1){G[a+n].push_back(b);G[b+n].push_back(a);}else {G[a].push_back(a+n);G[b].push_back(b+n);}}else if(op[0]=='X'){if(c==1){G[a+n].push_back(b);G[b+n].push_back(a);G[a].push_back(b+n);G[b].push_back(a+n);}else {G[a].push_back(b);G[b].push_back(a);G[a+n].push_back(b+n);G[b+n].push_back(a+n);}}}}void tarjan(int u,int fa){int v;low[u]=dfn[u]=++dfs_clock;Instack[u]=true;s.push(u);for(int i=0;i<G[u].size();i++){v=G[u][i];if(!dfn[v]){tarjan(v,u);low[u]=min(low[v],low[u]);}else if(Instack[v])low[u]=min(low[u],dfn[v]);}if(low[u]==dfn[u]){++scc_cnt;for(;;){v=s.top();s.pop();Instack[v]=false;sccno[v]=scc_cnt;if(v==u) break;}}}void find(int l,int r){memset(low,0,sizeof(low));memset(dfn,0,sizeof(dfn));memset(sccno,0,sizeof(sccno));memset(Instack,false,sizeof(Instack));scc_cnt=dfs_clock=0;for(int i=l;i<=r;i++)if(!dfn[i]) tarjan(i,-1);}void solve(){for(int i=0;i<n;i++){if(sccno[i]==sccno[i+n]){printf("NO\n");return ;}}printf("YES\n");}int main(){while(scanf("%d%d",&n,&m)!=EOF){init();getmap();find(0,2*n-1);solve();}return 0;}