BZOJ 3754 Tree之最小方差树 解题报告

<span style="font-family: arial, verdana, helvetica, sans-serif; font-size: 18px; background-color: rgb(228, 240, 248);">题目大意就是要求一个图的最小方差生成树</span>

<span style="font-family: arial, verdana, helvetica, sans-serif; font-size: 18px; background-color: rgb(228, 240, 248);">点数<=100, 边数<=2000, 1<=边权<=100且为整数</span>

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My code:

#include <cmath>#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>using namespace std;#define N 100 + 5#define M 2000 + 5#define eps 1e-9 int n, m, Fa[N], W[M], Ord[M];double ave, ans = 1e9; struct Edge{    int u, v;    double c;    Edge(int _u = 0, int _v = 0, double _c = 0.0) {u = _u, v = _v, c = _c;}    bool operator < (const Edge a) const {return c < a.c;}}E[M]; inline int Find(int x){    return x == Fa[x] ? x : Fa[x] = Find(Fa[x]);} inline double MST(int s, int delta){    for (int i = 1; i <= n; i ++)        Fa[i] = i;    double sum = 0;    for (int i = s, cnt = 1; i && i <= m && cnt < n; i += delta)    {        int t = Ord[i], u = Find(E[t].u), v = Find(E[t].v);        if (u != v)        {            cnt ++;            sum += E[t].c;            Fa[u] = v;        }    }    return sum;} inline bool cmp(int u, int v){    return E[u].c < E[v].c && abs(E[u].c - E[v].c) > eps;} inline double Solve(int sum){    ave = (double) sum / (n - 1);    for (int i = 1; i <= m; i ++)        E[i].c = (ave - W[i]) * (ave - W[i]);    sort(Ord + 1, Ord + m + 1, cmp);    return sqrt(MST(1, 1) / (n - 1));} int main(){    scanf("%d%d", &n, &m);    for (int i = 1; i <= m; i ++)    {        int u, v;        scanf("%d%d%d", &u, &v, W + i);        E[i] = Edge(u, v, (double) W[i]);        Ord[i] = i;    }    sort(Ord + 1, Ord + m + 1, cmp);    int Min = (int) MST(1, 1);    int Max = (int) MST(m, -1);    for (int i = Min; i <= Max; i ++)    {       double t = Solve(i);       ans = min(ans, t);    }    printf("%.4lf\n", ans);         return 0;}