Poj 3678 2-Sat

2-create a typical sat graph. Note that when "and" is 1, "a = 0" must be connected to "A = 1" and "B", because at this time, neither a nor B can be 0.

When or is 0, a = 1 must be connected to a = 0, and B is the same, because at this time, A and B cannot be 1.

#include<iostream>#include<cstdio>#include<cstring>#include<vector>#include<algorithm>#include<cmath>using namespace std;#define maxn 1005vector<int> g[maxn*2];bool mark[maxn*2];int S[maxn*2],cnt,n,m;void init(){    for(int i=0;i<n*2;i++) g[i].clear();    memset(mark,false,sizeof(mark));}void add(int x,int xval,int y,int yval){    x=x*2+xval;    y=y*2+yval;    g[x].push_back(y);}bool dfs(int x){    if(mark[x^1]) return false;    if(mark[x]) return true;    mark[x]=true;    S[cnt++]=x;    for(int i=0;i<g[x].size();i++) if(!dfs(g[x][i])) return false;    return true;}bool solve(){    for(int i=0;i<n*2;i+=2)    {        if(!mark[i]&&!mark[i+1])        {            cnt=0;            if(!dfs(i))            {                while(cnt>0) mark[S[--cnt]]=false;                if(!dfs(i+1)) return false;            }        }    }    return true;}void debug(int k){    printf("%d:",k);    for(int i=0;i<g[k].size();i++)    {        printf(" %d",g[k][i]);    }    puts("");}int main(){    int a,b,c;    char str[6];    while(~scanf("%d%d",&n,&m))    {        init();        for(int i=1;i<=m;i++)        {            scanf("%d%d%d%s",&a,&b,&c,str);            if(str[0]=='A')            {                if(c)                    add(a,0,a,1),                    add(b,0,b,1);                else                    add(a,1,b,0),                    add(b,1,a,0);            }            else if(str[0]=='O')            {                if(c)                    add(a,0,b,1),                    add(b,0,a,1);                else                    add(a,1,a,0),                    add(b,1,b,0);            }            else if(str[0]=='X')            {                if(c)                    add(a,0,b,1),                    add(a,1,b,0),                    add(b,0,a,1),                    add(b,1,a,0);                else                    add(a,1,b,1),                    add(a,0,b,0),                    add(b,0,a,0),                    add(b,1,a,1);            }        }        printf("%s\n",solve()?"YES":"NO");    }    return 0;}/*3 30 1 1 XOR1 2 1 XOR2 0 1 XOR*/