hdu 2122

kruskal模板题；

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>

using namespace std;

int n, m;
struct Edge{
    int u, v, w;
    void addEdge(int n_u = 0, int n_v = 0, int n_w = 0){
        u = n_u;
        v = n_v;
        w = n_w;
    }
}edge;
vector<Edge> evec;
int p[1005], cnt;

bool cmp(const Edge &a, const Edge &b){
    return a.w < b.w;
}
void K_init(){
    cnt = 0;    //初始化cnt ,cnt记录建立了多少条边;
    for(int i = 0; i < n; ++i)
        p[i] = i;
    sort(evec.begin(), evec.end(), cmp);
}
int Find(int x){
    return x == p[x] ? x : p[x] = Find(p[x]); //路径压缩优化;
}
void Union(int x, int y){
    int xRoot = Find(x);
    int yRoot = Find(y);
    p[xRoot] = yRoot;
}
int Kruskal(){
    K_init();
    int ans = 0;
    for(int i = 0; i < m; ++i){
        if(Find(evec[i].u) != Find(evec[i].v)){
            ans += evec[i].w;
            Union(evec[i].u, evec[i].v);
            cnt++;
        }
    }
    return ans;
}

int main()
{
    while(scanf("%d%d", &n, &m) != EOF){
        int u, v, w;
        for(int i = 0; i < m; ++i){
            scanf("%d%d%d", &u, &v, &w);
            edge.addEdge(u, v, w);
            evec.push_back(edge);
        }
        int ans = Kruskal();
        if(cnt != n-1)
            printf("impossible\n");
        else
            printf("%d\n", ans);
        printf("\n");
        evec.clear();
    }
    return 0;
}

当然也可以不使用STL，这份模板可能理解上难度大一些，加上一些注释；

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>

using namespace std;

int n, m;
int u[10005], v[10005], w[10005], r[10005]; // r间接排序;
int p[1005], cnt;

bool cmp(const int a, const int b){
    return w[a] < w[b];
}
void K_init(){
    cnt = 0;    //初始化cnt ,cnt记录建立了多少条边;
    for(int i = 0; i < n; ++i)  //初始化并查集;
        p[i] = i;
    for(int i = 0; i < m; ++i)  //初始化边序号;
        r[i] = i;
    sort(r, r+m, cmp);    //间接排序;
}
int Find(int x){
    return x == p[x] ? x : p[x] = Find(p[x]);  //路径压缩优化;
}

int Kruskal(){
    K_init();
    int ans = 0;
    for(int i = 0; i < m; ++i){
        int e = r[i];
        int x = Find(u[e]);
        int y = Find(v[e]);
        if(x != y){         //如果在不同集合,合并;
            ans += w[e];  
            p[x] = y;
            cnt++;
        }
    }
    return ans;
}

int main()
{
    while(scanf("%d%d", &n, &m) != EOF){
        for(int i = 0; i < m; ++i){
            scanf("%d%d%d", &u[i], &v[i], &w[i]);
        }
        int ans = Kruskal();
        if(cnt != n-1)
            printf("impossible\n");
        else
            printf("%d\n", ans);
        printf("\n");
    }
    return 0;
}