Uva11987almost Union-find (and set delete node)

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Test instructions: N number (i.e. 1-n) and M operations:

1 means merging x and Y, 2 means moving x to the y set, and 3 for counting the number of elements in the X collection

1,3 say, the key is 2 operation, you can remove 2, delete the operation can find an additional number to replace X, so that there is a new collection, moved to the Y set to merge

#include <iostream> #include <cstdio> #include <algorithm> #include <cstring>using namespace  std;const int N = 100000;typedef Long long ll;int father[n + n + 10];int sum[n + n + 10],cnt[n + n + 10]; Sum of all elements within the collection and, CNT represents the number of int value[n + N + 10];    Represents the first I numeric int n,m,k;void init () {k = n + 1;        for (int i = 0; I <= N; i++) {father[i] = i;        Sum[i] = i;        Cnt[i] = 1;    Value[i] = i;    }}int Find (int x) {if (x = = Father[x]) return x; return father[x] = Find (Father[x]);}    void Union (int x, int y) {int fx = Find (x);    int fy = Find (y);        if (FX! = FY) {father[fx] = FY;        Sum[fy] + = Sum[fx];    Cnt[fy] + = Cnt[fx];    }}void Remove (int x) {/* int fx = Find (value[x]);    cnt[fx]--;    SUM[FX]-= x;    Value[x] = k;    int fy = Find (value[y]);    Father[k] = FY;    cnt[fy]++;    SUM[FY] + = x;    k++;    */INT FX = Find (value[x]);  cnt[fx]--;  Remove X from the parent node collection minus one sum[fx]-= x; Subtract x value[x from sum] = k;        Using k instead of X, a separate set is generated, and the parent node is k cnt[k] = 1;    SUM[K] = x;    Father[k] = k; k++;}        int main () {while (scanf ("%d%d", &n, &m)! = EOF) {k = n + 1;            for (int i = 0; I <= N; i++) {father[i] = i;            Sum[i] = i;            Cnt[i] = 1;        Value[i] = i;        } int order,p,q;            while (m--) {scanf ("%d", &order);                if (order = = 1) {scanf ("%d%d", &p, &q);            Union (Value[p], value[q]);                } else if (order = = 2) {scanf ("%d%d", &p, &q);                int FX = Find (value[p]);                int fy = Find (value[q]); if (FX! = FY)//if p and Q are originally in a collection, then no operation is necessary.                    But the less this on WA, do not understand why the operation is not {Remove (P);                Union (Value[p], value[q]); }} else if (order = = 3) {scanf ("%d", &AMP;P);                int FX = Find (value[p]);            printf ("%d%d\n", Cnt[fx], sum[fx]); }}} return 0;}

　　

Uva11987almost Union-find (and set delete node)