hdu1824 Let's go home--2-sat

原题链接：​​http://acm.hdu.edu.cn/showproblem.php?pid=1824​​

题意：n个队伍，每一个队（三人一队），或者队长留下或者其余两名队员同时留下；接下来m对编号，每一对队员，如果队员A留下，则队员B必须回家休息下，或者B留下，A回家。

分析：用hash来保存编号，对于一个队三个人，给队长单独的hash值，再给剩下的两人相同的hash值，这样问题就转化了，很简单。

#define _CRT_SECURE_NO_DEPRECATE 

#include<iostream>
#include<vector>
#include<cstring>
#include<queue>
#include<stack>
#include<algorithm>
#include<cmath>
#define INF 99999999
#define eps 0.0001
using namespace std;

int n, m;
int index, cnt;
int dfn[2005];
int low[2005];
int belong[2005];
int hash1[30005];
bool inStack[2005];
vector<int> vec[2005];
stack<int> s;

void tarjan(int u)
{
  low[u] = dfn[u] = ++index;
  inStack[u] = 1;
  s.push(u);
  for (int i = 0; i < vec[u].size(); i++)
  {
    int v = vec[u][i];
    if (!dfn[v])
    {
      tarjan(v);
      low[u] = min(low[u], low[v]);
    }
    else if (inStack[v])
      low[u] = min(low[u], dfn[v]);
  }

  if (low[u] == dfn[u])
  {
    cnt++;
    int p;
    do
    {
      p = s.top();
      s.pop();
      inStack[p] = 0;
      belong[p] = cnt;
    } while (p != u);
  }
}

void solve()
{
  for (int i = 0; i < 2 * n; i++)
    if (!dfn[i])
      tarjan(i);

  for (int i = 0; i < 2 * n; i = i + 2)
  {
    if (belong[i] == belong[i + 1])
    {
      printf("no\n");
      return;
    }
  }
  printf("yes\n");
}

int main()
{
  while (~scanf("%d%d", &n, &m))
  {
    //init
    index = cnt = 0;
    for (int i = 0; i < 2 * n; i++)
    {
      vec[i].clear();
      dfn[i] = 0;
      inStack[i] = 0;
    }

    int x, y, z;
    int t = 0;
    for (int i = 0; i < n; i++)
    {
      scanf("%d%d%d", &x, &y, &z);
      hash1[x] = t++;
      hash1[y] = hash1[z] = t++;
    }

    for (int i = 0; i < m; i++)
    {
      scanf("%d%d", &x, &y);
      vec[hash1[x]].push_back(hash1[y] ^ 1);
      vec[hash1[y]].push_back(hash1[x] ^ 1);
    }

    solve();
  }

  return 0;
}