[UVA][并查集變形] 11987 - Almost Union-Find
Problem A
Almost Union-Find
I hope you know the beautiful Union-Find structure. In this problem, you're to implement something similar, but not identical.
The data structure you need to write is also a collection of disjoint sets, supporting 3 operations:
1 p q
Union the sets containing p and q. If p and q are already in the same set, ignore this command.
2 p q
Move p to the set containing q. If p and q are already in the same set, ignore this command
3 p
Return the number of elements and the sum of elements in the set containing p.
Initially, the collection contains n sets: {1}, {2}, {3}, ..., {n}.
Input
There are several test cases. Each test case begins with a line containing two integers n and m (1<=n,m<=100,000), the number of integers, and the number of commands. Each of the next m lines contains a command. For every operation, 1<=p,q<=n. The input is terminated by end-of-file (EOF). The size of input file does not exceed 5MB.
Output
For each type-3 command, output 2 integers: the number of elements and the sum of elements.
Sample Input
5 7 1 1 2 2 3 4 1 3 5 3 4 2 4 1 3 4 3 3
Output for the Sample Input
3 12 3 7 2 8
Explanation
Initially: {1}, {2}, {3}, {4}, {5}
Collection after operation 1 1 2: {1,2}, {3}, {4}, {5}
Collection after operation 2 3 4: {1,2}, {3,4}, {5} (we omit the empty set that is produced when taking out 3 from {3})
Collection after operation 1 3 5: {1,2}, {3,4,5}
Collection after operation 2 4 1: {1,2,4}, {3,5}
當我們將操作 2 的 p 點抽出時, 賦予它成為新的節點編號, 打個新的映射即可 !
#include <stdio.h>
int p[200001], r[200001], mapped[200001];
long long s[200001];
void init(int n) {
static int i;
for(i = 0; i <= n; i++) {
p[i] = i, s[i] = 0, r[i] = 1;
mapped[i] = i;
}
}
int find(int x) {
return p[x] == x ? x : p[x]=find(p[x]);
}
void joint(int x, int y, int i, int j) {
if(x == y)
return;
s[x] += s[y];
r[x] += r[y];
s[y] = 0;
r[y] = 0;
p[y] = x;
}
int main() {
int n, m, op, i, j, t, x, y;
while(scanf("%d %d", &n, &m) == 2) {
init(n+m);
for(i = 0; i <= n; i++)
s[i] = i;
while(m--) {
scanf("%d %d", &op, &i);
if(op == 3) {
t = find(mapped[i]);
printf("%d %lld\n", r[t], s[t]);
} else if(op == 2) {
scanf("%d", &j);
x = find(mapped[i]);
y = find(mapped[j]);
if(x != y) {
s[x] -= i, r[x]--;
mapped[i] = ++n;
s[mapped[i]] = i;
r[mapped[i]] = 1;
joint(find(mapped[i]), find(mapped[j]), i, j);
}
} else {
scanf("%d", &j);
joint(find(mapped[i]), find(mapped[j]), i, j);
}
}
}
return 0;
}