UVA11987- Almost Union-Find

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本题比较难处理的是删除之后根节点的归属，所以要开2倍的数组，如

void init() {    for (int i = 1; i <= n; i++) {        f[i] = i + n;        f[i + n] = i + n;        sum[i + n] = i;         cnt[i + n] = 1;    }}

再开两个数组分别记录集合元素的个数，以及集合元素的和。

#include <iostream>#include <cstdio>#include <cstring>#define N 100005using namespace std;int f[N * 2], cnt[N * 2], sum[N * 2];int n, m, a, b, k;void init() {    for (int i = 1; i <= n; i++) {        f[i] = i + n;        f[i + n] = i + n;        sum[i + n] = i;         cnt[i + n] = 1;    }}int getFather(int x) {    return x == f[x] ? x : f[x] = getFather(f[x]);}void Delete(int a, int b) {    int pa = getFather(a);    int pb = getFather(b);    if (pa != pb) {        f[a] = getFather(pb);         sum[pa] -= a;        sum[pb] += a;        cnt[pa]--;         cnt[pb]++;    }}void Union(int a, int b) {    int pa = getFather(a);    int pb = getFather(b);    if (pa != pb) {        f[pa] = getFather(pb);         sum[pb] += sum[pa];         cnt[pb] += cnt[pa];     }}int main() {    while (scanf("%d %d", &n, &m) != EOF) {        init();         for (int i = 0; i < m; i++) {            scanf("%d", &k);             if (k == 1) {                scanf("%d %d", &a, &b);                Union(a, b);            }               else if (k == 2) {                scanf("%d %d", &a, &b);                Delete(a, b);             }             else if (k == 3) {                scanf("%d", &a);                 int pa = getFather(a);                 printf("%d %d\n", cnt[pa], sum[pa]);             }        }     }    return 0;}

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