Just for funzjfc 并查集操作

题目描述

When I boring , I'm crazy...So I design a BT game just for fun...It's easy...

输入描述

    First line is a interger T <=20 ;
T case follow ,each case begin with the interger N(N<=100000) , as the max number I will give .
Then the interger M(M<=10000) ,as the M lines follow:
One type of format as "A B" is the only instruction I did,it means let A and B in the same set.
All the interger not less than 0.

输出描述

    if they are already in the same set puts "nop",else printf the min number in this set.

样例输入

1
10
5
1 3
1 2
5 4
5 3
5 3

样例输出

1
1
4
1
nop

使用并查集，

做每次归并操作的时候，取2个集合中的最小根结点作为新集合的根节点，并输出该

结点的值。同时可以做路径压缩操作。

代码如下：

#include<stdio.h>

int set[100002],n,m,a,b,t;

void init()

{

    int i;

    for(i=1;i<=n;i++){

        set[i]=i;

    }

}

int find2(int x)

{

    int r=x,temp;

    while(set[x]!=x)

        x=set[x];

while(r!=x)

temp=set[r],set[r]=x,r=temp;

    return x;

}

void merge2(int a,int b)

{

int ra=find2(a);

int rb=find2(b);

if(ra==rb)

printf("nop\n");

    else if(ra<rb){

        set[rb]=ra;

printf("%d\n",ra);

}

    else{

        set[ra]=rb;

printf("%d\n",rb);}

}

int main()

{

    scanf("%d",&t);

    while(t--){

scanf("%d",&n);

        init();

scanf("%d",&m);

        while(m--){

            scanf("%d %d",&a,&b);

            merge2(a,b);

        }    

    }

    return 0;

}