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题目链接：http://acm.swust.edu.cn/problem/856/
There are N trees in a forest. At first, each tree contains only one node as its root. And each node is marked with a number.
You‘re asked to do the following two operations:
A X Y, you need to link X‘s root to Y as a direct child. If X and Y have already been in the same tree, ignore this operation.
B X, you need to output the maximum mark in the chain from X to its root (inclusively).
The first line contains an integer T, indicating the number of followed cases. (1 <= T <= 20)
For each case, the first line contains two integers N and M, indicating the number of trees at beginning, and the number of operations follows, respectively. (1 <= N, M <= 100,000)
And the following line contains N integers, which are the marks of the N trees. (0 <= Mark <= 100,000)
And the rest lines contain the operations, in format A X Y, or B X, (0 <= X, Y < N).
For each ‘B X‘ operation, output the maximum mark.
|
1
2
3
4
5
6
7
8
9
|
1
5 5
5 4 2 9 1
A 1 2
A 0 4
B 4
A 1 0
B 1
|
|
1
2
3
|
1
5
|
题目大意：就是一个数组（下标重零开始），有对应的A,B操作，A a,b,是把a所在集合归并到b上，
若某一个集合已合并不进行操作，然B a,就是查询a集合中的最大值。
解题思路：运用并查集就是，注意a集合到b所在集合，并查集合并区分一下就可以了
代码如下：
1 #include <stdio.h> 2 int n, m, maxn, t, f[100005], cur[100005]; 3 4 void init(){ 5 scanf("%d%d", &n, &m); 6 for (int i = 0; i <= n; i++){ 7 scanf("%d", &cur[i]); 8 f[i] = i; 9 } 10 } 11 12 int findset(int x){ 13 maxn = cur[x]; 14 if (f[x] == x) return x; 15 int y = findset(f[x]); 16 if (maxn > cur[x]) cur[x] = maxn; 17 else maxn = cur[x]; 18 return f[x] = y; 19 } 20 21 void mergy(){ 22 int i, x, y; 23 char k[3]; 24 for (i = 0; i < m; i++){ 25 scanf("%s", k); 26 if (k[0] == ‘A‘){ 27 scanf("%d%d", &x, &y); 28 int a = findset(x), b = findset(y); 29 if (a != b) f[a] = y;//注意和传统并查集的区别 30 } 31 else{ 32 scanf("%d", &x); 33 findset(x); 34 printf("%d\n", cur[x]); 35 } 36 } 37 } 38 39 int main(){ 40 scanf("%d", &t); 41 while (t--){ 42 init(); 43 mergy(); 44 } 45 return 0; 46 }
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原文地址：http://www.cnblogs.com/zYx-ac/p/4564518.html
