一、思路
枚举所有生成树的边权和值,对每一个枚举的边权和值sum,修改所有边的边权为(es[i].cost - sum * 1.0 / (N - 1))2,即方差公式的分子,然后跑最小生成树算法,同时记录边的原来的权值和,如果求出的“最小方差”生成树的边权值和为sum,那么,用这个"最小方差"去更新答案。
二、复杂度分析
时间复杂度:O(N * W * M * logM)。N * W为枚举边权和值的时间。边权和值最小为0,最大为(N - 1) * W。
三、PS
这题据说蓝桥杯官网数据有问题。有5个样例(2、3、4、5、6),总是WA。所以,正确的代码也只能得50分。
四、源代码
#include<bits/stdc++.h> using namespace std; int N, M; const int MAXN = 55, MAXM = 1010; const double INF = (1LL << 60) * 1.0; typedef struct Edge0 { int u, v, oldcost; double newcost; bool operator < (Edge0 e) const { return newcost < e.newcost; } void assgin(int _u, int _v, int _cost) { u = _u; v = _v; oldcost = _cost; } } Edge; Edge edges[MAXM]; template <class T> inline void read(T &x) { int t; bool flag = false; while((t = getchar()) != ‘-‘ && (t < ‘0‘ || t > ‘9‘)) ; if(t == ‘-‘) flag = true, t = getchar(); x = t - ‘0‘; while((t = getchar()) >= ‘0‘ && t <= ‘9‘) x = x * 10 + t - ‘0‘; if(flag) x = -x; } /**并查集部分*/ int par[MAXN], rank[MAXN]; void init_ufind() { for(int i = 0; i < MAXN; ++i) { par[i] = i; rank[i] = 0; } } int ufind(int x) { return x == par[x] ? x : par[x] = ufind(par[x]); } void unite(int x, int y) { x = ufind(x), y = ufind(y); if(x == y)return; if(rank[x] < rank[y])par[x] = y; else { par[y] = x; if(rank[x] == rank[y])rank[x]++; } } bool same(int x, int y) { return ufind(x) == ufind(y); } /**并查集部分*/ double kruscal(int tot) { double avg = tot * 1.0 / (N - 1); for(int i = 0; i < M; ++i) { edges[i].newcost = (edges[i].oldcost * 1.0 - avg) * (edges[i].oldcost * 1.0 - avg); } sort(edges, edges + M); init_ufind(); double res = 0; int ires = 0; for(int i = 0; i < M; ++i) { Edge& e = edges[i]; if(!same(e.u, e.v)) { unite(e.u, e.v); res += e.newcost; ires += e.oldcost; } } if(ires == tot)return res / (N - 1); else return INF; } int main() { #ifndef ONLINE_JUDGE freopen("input.txt", "r", stdin); #endif int T = 1, a, b, c; int costs[MAXM]; while(scanf("%d%d", &N, &M), !(N == 0 && M == 0)) { for(int i = 0; i < M; ++i) { read(a), read(b), read(c); edges[i].u = a, edges[i].v = b, edges[i].oldcost = c; costs[i] = c; } sort(costs, costs + M); int mintot = 0, maxtot = 0; for(int i = 0;i < N - 1;++i)mintot += costs[i]; for(int i = M - 1;i > M - N;--i)maxtot += costs[i]; double ans = INF; for(int tot = mintot; tot <= maxtot; ++tot) { ans = min(ans, kruscal(tot)); } printf("Case %d: %.2f\n", T++, ans); } return 0; }
原文地址:https://www.cnblogs.com/565261641-fzh/p/8284533.html
时间: 2024-11-05 20:42:01