Just for fun----zjfc 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;


}