HDU 2473 Junk-Mail Filter （删点并查集）

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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;}

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