Kruskal disjoint-set minimum spanning tree

Title Description
As stated, given an undirected graph, the minimum spanning tree is obtained, if this figure is not connected, the output orz

Input Output Format
Input format:
The first line contains two integers N, M, N represents the total FIG nodes and the M undirected edges. (N <= 5000, M < = 200000)
Next M lines contains three integers Xi, Yi, Zi, Zi expressed a length of edges connected to node No Xi, Yi

Output format:
output contains a number, i.e., the minimum spanning tree and the length of each side; if the output is not connected orz FIG.

Sample Input Output
Input Sample # 1:
. 4. 5
1 2 2
1. 3 2
1. 4. 3
2. 3. 4
. 3. 4. 3
Output Sample # 1:
7

This code actually appeared a little mistake, but I really do not know what wrong. . .
Well, I later modified the merge code, be written to a

#include<iostream>
#include<stdio.h>
#include<algorithm>
using namespace std;
//最小生成树之kruskal 也就是使用并查集
//第一个是判断 边是否在同一个子集之中 第二个就是判断 边之间的合并
int a[100]={0};//
//再创建一个结构体数组来存储 起点，终点 和边的长度
struct jj{
    int x,y,z;
}J[100];
//写一个判断祖宗的方法
int zuzong(int x){
    if(a[x]==x)
        return x;
    else
        return zuzong(a[x]);
}
//写一个判断的方法
bool check(int x,int y){
   return zuzong(x)==zuzong(y);
}

//再写一个合并的方法
void merge_1(int x,int y){
    if(!check(x,y))
    a[zuzong(x)]=zuzong(y);

}
//再写一个比较的方法,用边的长度作为比较对象
bool cmp(jj x,jj y){//名字要取好
    return x.z<y.z;
}
int main()
{
    int N,M;//有n个结点，M条边
    int sum=0;
    int co=0;//这个是记录边数目，也就是能不能生成最小生成树
    int i;
    //先对结点和边进行输入
    scanf("%d%d",&N,&M);
    for(i=0;i<M;i++){
        scanf("%d%d%d",&J[i].x,&J[i].y,&J[i].z);
    }
    //然后按照思想，先对边进行从小到大排序，然后使用并查集对边的顶点进行判断是否在同一个集合中
    sort(J,J+M,cmp);
    //对几个结点的祖宗赋初值,都是等于自己
    for(i=1;i<=N;i++)
        a[i]=i;

    //然后开始判断
    for(i=0;i<M;i++){//从第一条边到最后一条边
        if(!check(J[i].x,J[i].y)){//如果这条边的两个点不在同一个集合，那就合并，并且记录长度
            merge_1(J[i].x,J[i].y);
            sum+=J[i].z;
            co++;
        }
    }

    //最后输出
    if(co==N-1)
        printf("%d",sum);
    else
        printf("orz");

}