#operator ——“Kruskal算法求最小生成树 中的 operator” ~20.8.17

这里是找到的资料
mycomplex operator + (mycomplex &p1,mycomplex &p2) //重定义"+"
{
	 mycomplex p;
	 p.x=p1.x+p2.x;
	 p.y=p2.y+p1.y;
	 return p;
 }

int main()
{
    mycomplex p1（8，2）,p2（8，1）;
    mycomplex p3;
    p3=p1+p2;
    std::cout<<p3<<std::endl;
    return 0;
}

原题代码
#include<bits/stdc++.h>
using namespace std;
const int N = 200010;

int n, m;
int p[N];

struct E{
    int a, b, w;
    bool operator < (E const &W) const{//这里
        return w < W.w;
    }
}e[N];

int find(int x){
    if(p[x] != x) p[x] = find(p[x]);
    return p[x];
}

int main(){
    cin >> n >> m;
    for(int i = 0; i <  m; i ++){
        int a, b, w;
        cin >> a >> b >> w;
        e[i] = {a, b, w};
    }
    
    sort(e, e + m);
    
    for(int i = 1; i < n; i ++) p[i] = i;
    
    int res = 0, cnt = 0;
    for(int i = 0 ; i < m; i ++){
        int a = e[i].a, b = e[i].b, w = e[i].w;
        
        a = find(a), b = find(b);
        if(a != b){
            p[a] = b;
            res += w;
            cnt ++;
        }
    }
    if(cnt < n - 1) cout << "impossible";
    else cout << res;
    
    return 0;
}

具体笔记
bool operator < (E const &W) const{
//这里 可以写成 bool operator < (const E & W) const{
        return w < W.w;
    }
/*
*1 bool operator                是    重定义 < 时返回的类型。
*2 (E const &W) / (const E & W) 是    E中所有的元素W
*3 return w < W.w;              是    排序的标准(从小到大)
*/

建议用以下写法

struct E{
    int a, b, w;
}e[N];
//像这样将operator放到外面去，好理解…… 
bool operator < (E &a, E &b){
    return a.w < b.w;
}

#include<bits/stdc++.h>
using namespace std;
const int N = 200010;

int n, m;
int p[N];
//定义数据结构E,方便对边进行操作
struct E{
    int a, b, w;
}e[N];

bool operator < (E &a, E &b){
    return a.w < b.w;
}

int find(int x){
    if(p[x] != x) p[x] = find(p[x]);
    return p[x];
}

int main(){
    cin >> n >> m;
    for(int i = 0; i < m; i ++){
        int a, b, w;
        cin >> a >> b >> w;
        e[i] = {a, b, w};
    }
    
    sort(e, e + m);
    
    for(int i = 1; i < n; i ++) p[i] = i;
    
    int res = 0, cnt = 0;
    for(int i = 0; i < m; i ++){
        int a = e[i].a, b = e[i].b, w = e[i].w;
        
        a = find(a), b = find(b);
        if(a != b){
            p[a] = b;;
            res += w;
            cnt ++;
        }
    }
    
    if(cnt < n - 1) cout << "impossible";
    else cout << res;
    
    return 0;
}