[Swust OJ 856]--Huge Tree(并查集)

题目链接：http://acm.swust.edu.cn/problem/856/

Time limit(ms): 1000　　　　　　　　Memory limit(kb): 10000

Description

　　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).

Input

　　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).

Output

　　For each 'B X' operation, output the maximum mark.

Sample Input

　　

1

5 5

5 4 2 9 1

A 1 2

A 0 4

B 4

A 1 0

B 1

 

Sample Output

 

　　

1

5

 

 

题目大意：就是一个数组（下标从零开始），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：

#include <stdio.h>
int n, m, maxn, t, f[100005], cur[100005];

void init(){
    scanf("%d%d", &n, &m);
    for (int i = 0; i <= n; i++){
        scanf("%d", &cur[i]);
        f[i] = i;
    }
}

int findset(int x){
    maxn = cur[x];
    if (f[x] == x) return x;
    int y = findset(f[x]);
    if (maxn > cur[x]) cur[x] = maxn;
    else maxn = cur[x];
    return f[x] = y;
}

void mergy(){
    int i, x, y;
    char k[3];
    for (i = 0; i < m; i++){
        scanf("%s", k);
        if (k[0] == 'A'){
            scanf("%d%d", &x, &y);
            int a = findset(x), b = findset(y);
            if (a != b) f[a] = y;//注意和传统并查集的区别
        }
        else{
            scanf("%d", &x);
            findset(x);
            printf("%d
", cur[x]);
        }
    }
}

int main(){
    scanf("%d", &t);
    while (t--){
        init();
        mergy();
    }
    return 0;
}