[AcWing 859] Kruskal算法求最小生成树
作者:互联网
复杂度 \(O(m*log(m))\)
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#include<iostream>
#include<algorithm>
using namespace std;
const int N = 1e5 + 10, M = 2 * N, INF = 0x3f3f3f3f;
int n, m;
int p[N];
int res, cnt;
struct Edge {
int a, b, w;
bool operator< (const Edge & W) const {
return w < W.w;
}
}edges[M];
int find(int x)
{
if (p[x] != x) p[x] = find(p[x]);
return p[x];
}
int Kruskal()
{
sort(edges, edges + m);
for (int i = 1; i <= n; i ++) p[i] = i;
for (int i = 0; i < m; i ++) {
int a = edges[i].a, b = edges[i].b, w = edges[i].w;
a = find(a), b = find(b);
if (a != b) {
p[a] = b;
res += w;
cnt ++;
}
}
if (cnt < n - 1) return INF;
else return res;
}
int main()
{
cin >> n >> m;
for (int i = 0; i < m; i ++) {
int a, b, w;
cin >> a >> b >> w;
edges[i] = {a, b, w};
}
int t = Kruskal();
if (t == INF) puts("impossible");
else cout << res << endl;
return 0;
}
- Kruskal 思路:每次取出权值最小的边,如果边的两个端点之间没有连通,则把两个端点放到同一个并查集当中;
- 要先对所有的边进行排序,使用 sort 函数进行排序,在 struct 中对小于号进行重载;
标签:const,int,Kruskal,859,edges,include,find,AcWing 来源: https://www.cnblogs.com/wKingYu/p/16242717.html