题目

https://www.luogu.com.cn/problem/P1195

 

 分析

本题运用了一个贪心的思想，连接一条边就相当于连通块减1，运用kruskal算法的思想：每次连可以连的边中代价最小的 （贪心） 使用 并查集维护

代码

#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include<algorithm>
using namespace std;
#define maxn 10010
#define maxm 10010
struct node
{
    int from;
    int to;
    int dis;
}e[maxm * 2];
int edges[maxn][maxn];
int n, m, k;
int father[maxn];
bool cmp(struct node &a, struct node &b)
{
    return a.dis < b.dis;
}
int find(int x)
{
    if (father[x] == x)return x;
    return father[x] = find(father[x]);
}
int allcount = 0;
int counts = 0;
void kruskal()
{
        sort(e+1, e + m+1, cmp);
    for (int i = 1; i <=m; i++)
    {
        int tempx = find(e[i].from);
        int tempy = find(e[i].to);
        if (tempx == tempy)continue;
        counts++;
        father[tempx] = tempy;
        allcount += e[i].dis;
        if (counts== n - k) { printf("%d", allcount); return; }
    }
    printf("No Answer\n");
    return;
}

int main()
{
    scanf("%d%d%d", &n, &m, &k);
    for (int i = 1; i <=m; i++)
    {
        scanf("%d%d%d", &e[i].from, &e[i].to, &e[i].dis);
    }
    for (int i = 0; i <= n; i++)father[i] = i;
    kruskal();

}