蓝桥杯练习题 最小方差生成树 (Kruskal MST 好题)

1<=U,V<=N<=50,N-1<=M<=1000,0<=W<=50。数据不超过5组。

题目分析：要求方差最小，就是要每条边(val - ave)^2的和最小，枚举所有边权和的可能值，多次kruskal求最小生成树，每次求的时候，以(val - ave)^2作为当前边的权值，如果该树的val和等于我们枚举的和，则修改ans的值，因为题目的数据量很小，复杂度大概为O(NWElogE)大概是1e7左右，基本可以接受

#include <cstdio>
#include <algorithm>
using namespace std;double const MAX = 10000000000000.0;
int n, m, tmp[1005], fa[55];
double ans;struct Edge
{int u, v;double w, val;
}e[1005];bool cmp(Edge a, Edge b)
{return a.w < b.w;
}void UF_set(int n)
{for(int i = 1; i <= n; i++)fa[i] = i;
}int Find(int x)
{return x == fa[x] ? x : fa[x] = Find(fa[x]);
}void Union(int a, int b)
{int r1 = Find(a);int r2 = Find(b);if(r1 != r2)fa[r2] = r1;
}void Kruskal(int sum)
{UF_set(n);int cnt = 0;double f_all = 0;double all = 0;double ave = sum * 1.0 / (n - 1);for(int i = 0; i < m; i++)e[i].w = (e[i].val - ave) * (e[i].val - ave);sort(e, e + m, cmp);for(int i = 0; i < m; i++){int u = e[i].u;int v = e[i].v;if(Find(u) != Find(v)){Union(u, v);f_all += e[i].w;all += e[i].val;cnt ++;}if(cnt == n - 1)break;}if((int)all == sum)ans = min(ans, f_all);
}int main()
{int ca = 1;while(scanf("%d %d", &n, &m) != EOF && (m + n)){// if(n == 1 || n == 2)// {//     printf("0.00\n");//     continue;// }int minv = 0;int maxv = 0;ans = MAX;for(int i = 0; i < m; i++){scanf("%d %d %lf", &e[i].u, &e[i].v, &e[i].val);tmp[i] = e[i].val;}sort(tmp, tmp + m);for(int i = 0; i < n - 1; i++)minv += tmp[i];for(int i = m - 1; i > m - n; i--)maxv += tmp[i];for(int i = minv; i <= maxv; i++)Kruskal(i);ans = ans / (n - 1);printf("Case %d: %.2f\n", ca++, ans);}
}