[swustoj 856] Huge Tree

Huge Tree(0856)

问题描述

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).

输入

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).

输出

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

样例输入

1
5 5
5 4 2 9 1
A 1 2
A 0 4
B 4
A 1 0
B 1 

样例输出

1
5

简单并查集、

#include<iostream>

#include<cstdio>

#include<cstring>

#include<algorithm>

using namespace std;

#define N 100010



int n,m;

int val[N];

int mx[N];

int f[N];



void init()

{

    for(int i=1;i<=n;i++){

        f[i]=i;

        mx[i]=val[i];

    }

}

int Find(int x)

{

    if(x==f[x]) return x;

    int t=f[x];

    f[x]=Find(t);

    mx[x]=max(mx[x],mx[t]);

    return f[x];

}

void UN(int x,int y)

{

    x=Find(x);

    //y=Find(y);

    if(x==y) return;

    f[x]=y;

}

int main()

{

    int T;

    scanf("%d",&T);

    while(T--)

    {

        scanf("%d%d",&n,&m);

        for(int i=1;i<=n;i++) scanf("%d",&val[i]);

        init();

        while(m--){

            char op;

            int a,b;

            scanf(" %c",&op);

            if(op=='A'){

                scanf("%d%d",&a,&b);

                a++;

                b++;

                UN(a,b);

            }

            else{

                scanf("%d",&a);

                a++;

                Find(a);

                printf("%d\n",mx[a]);

            }

        }

    }

    return 0;

}