并查集入门

先贴上参考资料

1、杭电的并查集课件...

http://wenku.baidu.com/link?url=IJ10qA1w6APfrbZuF-tq0Rl_-SAM4_LyXBLpkAn_GUwlBhaFRZfMZN_XdUlXkwIhRW1SCp3tJBkqtaU7kOzOq33ENgByKAgHA3jtkKzTQ97

2、相关博客...

http://hi.baidu.com/czyuan_acm/item/13cbd3258c29e20d72863edf

http://www.cppblog.com/yuan1028/archive/2011/02/13/139990.html

---

并查集，顾名思义，就是做 “并” 和 “查” 两件事的。我们用到并查集基本都是模拟树形结构的，线形的效率太低。

先贴上模版

#include<cstdio>
#define MAXN 50005
int father[MAXN], rank[MAXN];
void init(int n)
{
  for (int i = 1; i <= n; ++i)
  {
    father[i] = i;
    rank[i] = 0;
  }
}
int find(int x)
{
  if (father[x] != x)
  {
    father[x] = find(father[x]);
  }
  return father[x];
}
void merge(int a, int b)
{
  int fa = find(a);
  int fb = find(b);
  if (rank[fa] < rank[fb])
  {
    father[fa] = fb;
  }
  if (rank[a] > rank[fb])
  {
    father[fb] = fa;
  }
  if (rank[fa] == rank[fb])
  {
    father[fb] = fa;
    rank[fa]++;
  }
}

1、father[]用来记录此结点的父结点，rank[]用来记录此树的深度

2、find()函数中用了路径压缩，在寻找某个结点的祖先时，将这条路径上所有结点的父结点更新为祖先结点

3、merge()函数中，是按秩合并。秩就是树的高度。不过很多时候没有必要按秩合并，也就不用开rank[]，直接下面这样就ok

void merge(int a, int b)
{
  int fa = find(a);
  int fb = find(b);
  father[fa] = fb;
}

---

练习题：

1、HDOJ 1213 How Many Tables(水题，直接套模版)

http://acm.hdu.edu.cn/showproblem.php?pid=1213

2、HDOJ 1232 畅通工程(同上、、)

http://acm.hdu.edu.cn/showproblem.php?pid=1232

3、POJ 1182 食物链(经典问题，里面的公式比较难想、、)

http://poj.org/problem?id=1182

下面是AC代码：

#include<cstdio>
#define MAXN 1005
int father[MAXN];
void init()
{
  for (int i = 0; i < MAXN; ++i)
  {
    father[i] = i;
  }
}
int find(int n)
{
  if (father[n] != n)
  {
    father[n] = find(father[n]);
  }
  return father[n];
}
void merge(int a, int b)
{
  if (find(a) != find(b))
  {
    father[find(b)] = find(a);
  }
}
int main()
{
  int T;
  scanf("%d", &T);
  while(T--)
  {
    int n, m, a, b, count = 0;
    init();
    scanf("%d %d", &n, &m);
    for (int i = 0; i < m; ++i)
    {
      scanf("%d %d", &a, &b);
      merge(a, b);
    }
    for (int i = 1; i <= n; ++i)
    {
      if (father[i] == i)
      {
        count++;
      }
    }
    printf("%d\n", count);
  }
  return 0;
}

#include<cstdio>
#include<cstdlib>
#define MAXN 1000
int father[MAXN];
void init()
{
  for (int i = 0; i < MAXN; ++i)
  {
    father[i] = i;
  }
}
int find(int n)
{
  int r = n;
  while(father[r] != r)
  {
    r = father[r];
  }
  return r;
}
void merge(int s, int e)
{
  int a = find(s);
  int b = find(e);
  if (a < b)
  {
    father[b] = a;
  }else
  {
    father[a] = b;
  }
}

int main()
{
  int n, m, s, e;
  while(scanf("%d", &n) != EOF)
  {
    int ans = 0;
    if (n == 0)
    {
      exit(0);
    }else
    {
      init();
      scanf("%d", &m);
      for (int i = 0; i < m; ++i)
      {
        scanf("%d %d", &s, &e);
        merge(s, e);
      }
      for (int i = 1; i <= n; ++i)
      {
        if (find(i) == i)
        {
          ans += 1;
        }
      }
      printf("%d\n", ans-1);
    }
  }
  return 0;
}

#include <cstdio>
#include <cstring>
using namespace std;
#define MAX 50050
int par[MAX],rel[MAX];

void init(int n)
{
    for(int i=1;i<=n;i++)
    {
        par[i]=i;
        rel[i]=0;
    }
}

int find(int x)
{
    if(par[x]==x)    return x;
    int tmp=par[x];
    par[x]=find(tmp);
    rel[x]=(rel[tmp]+rel[x])%3;
    return par[x];
}

void union_set(int x,int y,int px,int py,int d)
{
    par[px]=py;
    rel[px]=(rel[y]-rel[x]+2+d)%3;
}

int main()
{
    int T,n,m,a,b,pa,pb,k,r;
    scanf("%d%d",&n,&m);
    {
        init(n);
        r=0;
        for(int i=0;i<m;i++)
        {
            scanf("%d%d%d",&k,&a,&b);
            if(a>n||b>n)    {    r++;    continue;}
            if(k==2&&a==b)    {    r++;    continue;}
            pa=find(a);
            pb=find(b);
            if(pa==pb)
            {
                if((rel[b]+k+2)%3!=rel[a])    r++;
            }
            else    union_set(a,b,pa,pb,k);
        }
        printf("%d\n",r);
    }
}