HDU 2473 Junk-Mail Filter （删点并查集）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2473

题意：给n个点，m个操作，操作分两种，可以把两点合在同一集合或者把某点从原集合分离出去

思路：并查集要增添一个“删除”功能，即保持原树的形态不变，新建一个节点，把要删除的点的信息映射到该点上，之后对此点操作即可

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <queue>
#include <vector>
#include <utility>
#include <functional>
#include <string>
#include <set>

using namespace std;

const int maxn = 2000010;
const int mod = 1e9 + 7;
const int inf = 0x3f3f3f3f;

int fa[maxn], id[maxn];

void init(int n)
{
for (int i = 0; i < n; i++)
fa[i] = id[i] = i;
}

int find(int x)
{
return x == fa[x] ? x : fa[x] = find(fa[x]);
}

void unite(int x, int y)
{
x = find(x), y = find(y);
if (x != y)
fa[x] = y;
}

int main()
{
int n, m, ca = 1;
while (~scanf("%d%d", &n, &m) && (n + m))
{
init(n);
int cnt = n;
while (m--)
{
char c;
scanf(" %c", &c);
if (c == 'M')
{
int x, y;
scanf("%d%d", &x, &y);
unite(id[x], id[y]);
}
else
{
int x;
scanf("%d", &x);
id[x] = cnt;
fa[cnt] = cnt;
cnt++;
}
}

set <int> s;
for (int i = 0; i < n; i++)
s.insert(find(id[i]));
printf("Case #%d: %d\n", ca++, s.size());
}
return 0;
}