本文主要是介绍POJ 3925 Minimal Ratio Tree 最小生成树，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
思路是枚举+最小生成树，用DFS枚举或者二进制枚举。
/*
ID: sdj22251
PROG: calfflac
LANG: C++
*/
#include <iostream>
#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <cmath>
#include <ctime>
#define MAX 100000000
#define LOCA
#define PI acos(-1.0)
using namespace std;
int d[16][16], node[16], v[16], dis[16], ans[16];
int n, m;
double mi;
bool used[16];
void search()
{int i, j, k, sum, weight;sum = 0;weight = 0;memset(used, false, sizeof(used));for(i = 0; i < m; i++)dis[v[i]] = MAX;for(i = 0; i < m; i++)dis[v[i]] = d[v[0]][v[i]];used[v[0]] = true;for(i = 1; i < m; i++){int mini = MAX;int tmp = -1;for(j = 0; j < m; j++)if(!used[v[j]] && dis[v[j]] < mini){mini = dis[v[j]];tmp = v[j];}used[tmp] = true;sum += mini;for(j = 0; j < m; j++){if(!used[v[j]] && d[v[j]][tmp] < dis[v[j]])dis[v[j]] = d[v[j]][tmp];}}for(j = 0; j < m; j++)weight += node[v[j]];double T = (double)sum / (double)weight;if(T < mi){for(j = 0; j < m; j++)ans[j] = v[j];mi = T;}
}
void dfs(int x, int cnt)
{v[cnt] = x;if(cnt == m - 1){search();return;}for(int i = x + 1; i <= n; i++)dfs(i, cnt + 1);
}
int main()
{
#ifdef LOCALfreopen("ride.in","r",stdin);freopen("ride.out","w",stdout);
#endifint i, j;while(scanf("%d%d", &n, &m) != EOF){if(n == 0 && m == 0) break;for(i = 1; i <= n; i++)scanf("%d", &node[i]);for(i = 1; i <= n; i++){for(j = 1; j <= n; j++){scanf("%d", &d[i][j]);}}mi = 100000000.0;for(i = 1; i <= n; i++)dfs(i, 0);for(i = 0; i < m - 1; i++)printf("%d ", ans[i]);printf("%d\n", ans[m - 1]);}return 0;
}
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