UVA 11987 Almost Union-Find

并查集。三种操作，1 p q 表示把p所在集合都移到q所在集合里，2 p q 表示把p这一个点移到q所在集合里，3 q 要求输出q所在集合的元素个数及总和。

这道题的难点应该是2号操作，如何只移动一个节点呢？其实数据量其实不大，完全可以开两倍的数组。不妨假设一开始时i节点的父节点为n+i，这样就有n个虚节点，即使要把某棵树的根节点移走，也会留下虚节点，处理就会简单很多。然后在用两个数组记录对应的元素个数及总和。

这几乎就是一个并查集。

#include<algorithm>#include<iostream>#include<cstring>#include<cstdio>#include<vector>#include<string>#include<queue>#include<cmath>///LOOP#define REP(i, n) for(int i = 0; i < n; i++)#define FF(i, a, b) for(int i = a; i < b; i++)#define FFF(i, a, b) for(int i = a; i <= b; i++)#define FD(i, a, b) for(int i = a - 1; i >= b; i--)#define FDD(i, a, b) for(int i = a; i >= b; i--)///INPUT#define RI(n) scanf("%d", &n)#define RII(n, m) scanf("%d%d", &n, &m)#define RIII(n, m, k) scanf("%d%d%d", &n, &m, &k)#define RIV(n, m, k, p) scanf("%d%d%d%d", &n, &m, &k, &p)#define RV(n, m, k, p, q) scanf("%d%d%d%d%d", &n, &m, &k, &p, &q)#define RFI(n) scanf("%lf", &n)#define RFII(n, m) scanf("%lf%lf", &n, &m)#define RFIII(n, m, k) scanf("%lf%lf%lf", &n, &m, &k)#define RFIV(n, m, k, p) scanf("%lf%lf%lf%lf", &n, &m, &k, &p)#define RS(s) scanf("%s", s)///OUTPUT#define PN printf("\n")#define PI(n) printf("%d\n", n)#define PIS(n) printf("%d ", n)#define PS(s) printf("%s\n", s)#define PSS(s) printf("%s ", n)///OTHER#define PB(x) push_back(x)#define CLR(a, b) memset(a, b, sizeof(a))#define CPY(a, b) memcpy(a, b, sizeof(b))#define display(A, n, m) {REP(i, n){REP(j, m)PIS(A[i][j]);PN;}}using namespace std;typedef long long LL;typedef pair<int, int> P;const int MOD = 100000000;const int INFI = 1e9 * 2;const LL LINFI = 1e17;const double eps = 1e-6;const double pi = acos(-1.0);const int N = 222222;const int M = 22;const int move[8][2] = {0, 1, 0, -1, 1, 0, -1, 0, 1, 1, 1, -1, -1, 1, -1, -1};int n, k;int f[N], sum[N], num[N];int find(int x){return x == f[x] ? x : f[x] = find(f[x]);}void Union(int a, int b){    int x = find(a);    int y = find(b);    sum[y] += sum[a];    num[y] += num[a];    sum[x] -= sum[a];    num[x] -= num[a];    f[a] = y;}int main(){    //freopen("input.txt", "r", stdin);    //freopen("output.txt", "w", stdout);    int m, a, b, c, x, y;    while(RII(n, m) != EOF)    {        k = n + n;        FFF(i, 1, n)sum[i] = sum[n + i] = i, num[i] = num[i + n] = 1;        FFF(i, 1, n)f[i] = n + i;        FFF(i, n + 1, k)f[i] = i;        while(m--)        {            RI(c);            if(c == 3)            {                RI(a);                b = find(a);                PIS(num[b]), PI(sum[b]);            }            else            {                RII(a, b);                x = find(a);                y = find(b);                if(x != y)                {                    if(c == 1)Union(x, y);                    else Union(a, b);                }            }        }    }    return 0;}