HDU 1824 Let's go home (2-SAT)

思路：用op數組記錄隊員i的對立編號，如op[0]=1,op[1]=op[2]=0;,然後用2-SAT來判斷

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int maxn=3002;
const int maxm=12002;
int n,m,s;
struct node
{
    int to,next;
} e[maxm];
int head[maxn],edge;
inline void addedge(int u,int v)
{
    e[edge].to=v,e[edge].next=head[u],head[u]=edge++;
}
bool instack[maxn];
int dfn[maxn],low[maxn],stap[maxn],belong[maxn],cnt,ccnt,stop;
int op[maxn];
void init()
{
    stop=ccnt=cnt=edge=0;
    memset(head,-1,sizeof(head));
    memset(belong,0,sizeof(belong));
    memset(dfn,0,sizeof(dfn));
}
void dfs(int i)
{
    int v,u=i;
    dfn[u]=low[u]=++cnt;
    instack[i]=true;
    stap[stop++]=u;
    for(int j=head[u]; ~j; j=e[j].next)
    {
        if(!dfn[v=e[j].to])
        {
            dfs(v);
            low[u]=min(low[u],low[v]);
        }
        else if(instack[v])
            low[u]=min(low[u],dfn[v]);
    }
    if(low[u]==dfn[u])
    {
        ccnt++;
        while(true)
        {
            v=stap[--stop];
            belong[v]=ccnt;
            instack[v]=false;
            if(v==u) break;
        }
    }
}
void tarjan()
{
    for(int i=0; i<s; i++)
        if(!dfn[i])
            dfs(i);
}
bool ok()
{
    for(int i=0; i<s; i++)
        if(belong[i]==belong[op[i]]) return 1;
    return 0;
}
int main()
{
    int a,b,c;
    while(~scanf("%d%d",&n,&m))
    {
        init();
        s=3*n;
        while(n--)
        {
            scanf("%d%d%d",&a,&b,&c);
            addedge(b,c);
            addedge(c,b);
            op[a]=b;
            op[b]=op[c]=a;
        }
        while(m--)
        {
            scanf("%d%d",&a,&b);
            addedge(a,op[b]);
            addedge(b,op[a]);
        }
        tarjan();
        if(ok())  puts("no");
        else puts("yes");
    }
    return 0;
}