Description
For a tree, which nodes and edges are all weighted, the ratio of it is calculated according to the following equation. 
Input
Input contains multiple test cases. The first line of each test case contains two integers n (2<=n<=15) and m (2<=m<=n), which stands for the number of nodes in the graph and the number of nodes in the minimal ratio tree. Two zeros end the input. The next line contains n numbers which stand for the weight of each node. The following n lines contain a diagonally symmetrical n×n connectivity matrix with each element shows the weight of the edge connecting one node with another. Of course, the diagonal will be all 0, since there is no edge connecting a node with itself.
All the weights of both nodes and edges (except for the ones on the diagonal of the matrix) are integers and in the range of [1, 100].
The figure below illustrates the first test case in sample input. Node 1 and Node 3 form the minimal ratio tree.
All the weights of both nodes and edges (except for the ones on the diagonal of the matrix) are integers and in the range of [1, 100].
The figure below illustrates the first test case in sample input. Node 1 and Node 3 form the minimal ratio tree.
Output
For each test case output one line contains a sequence of the m nodes which constructs the minimal ratio tree. Nodes should be arranged in ascending order. If there are several such sequences, pick the one which has the smallest node number; if there's a tie, look at the second smallest node number, etc. Please note that the nodes are numbered from 1 .
题目大意：给n个点，一个完全图，要求你选出m个点和m-1条边组成一棵树，其中sum(边权)/sum(点权)最小，并且字典序最小，输出这m个点。
思路：大水题，n个选m个，$C_{n}^{m}$最大也不到1W，最小生成树算法也才$O(n^2)$，果断暴力。暴力枚举选和不选，然后用最小生成树求sum(边权)，逐个比较即可。
PS：太久没写最小生成树结果混入了最短路的东西结果WA了>_<
代码（15MS）：
1 #include <cstdio> 2 #include <cstring> 3 #include <iostream> 4 using namespace std; 5 6 const int MAXN = 18; 7 const int INF = 0x3fff3fff; 8 9 int mat[MAXN][MAXN]; 10 int weight[MAXN]; 11 int n, m; 12 bool use[MAXN], vis[MAXN]; 13 int dis[MAXN]; 14 15 int prim(int st) { 16 memset(vis, 0, sizeof(vis)); 17 vis[st] = true; 18 for(int i = 1; i <= n; ++i) dis[i] = mat[st][i]; 19 int ret = 0; 20 for(int cnt = 1; cnt < m; ++cnt) { 21 int u, min_dis = INF; 22 for(int i = 1; i <= n; ++i) 23 if(use[i] && !vis[i] && dis[i] < min_dis) u = i, min_dis = dis[i]; 24 ret += min_dis; 25 vis[u] = true; 26 for(int i = 1; i <= n; ++i) 27 if(use[i] && !vis[i] && dis[i] > mat[u][i]) dis[i] = mat[u][i]; 28 } 29 return ret; 30 } 31 32 bool ans[MAXN]; 33 int ans_pw, ans_ew; 34 35 void dfs(int dep, int cnt, int sum_pw) { 36 if(cnt == m) { 37 int sum_ew = prim(dep - 1); 38 if(ans_ew == INF || ans_ew * sum_pw > ans_pw * sum_ew) {//ans_ew/ans_pw > sum_ew/sum_pw 39 for(int i = 1; i <= n; ++i) ans[i] = use[i]; 40 ans_ew = sum_ew; ans_pw = sum_pw; 41 } 42 return ; 43 } 44 if(dep == n + 1) return ; 45 use[dep] = true; 46 dfs(dep + 1, cnt + 1, sum_pw + weight[dep]); 47 use[dep] = false; 48 dfs(dep + 1, cnt, sum_pw); 49 } 50 51 int main() { 52 while(scanf("%d%d", &n, &m) != EOF) { 53 if(n == 0 && m == 0) break; 54 for(int i = 1; i <= n; ++i) scanf("%d", &weight[i]); 55 for(int i = 1; i <= n; ++i) { 56 for(int j = 1; j <= n; ++j) scanf("%d", &mat[i][j]); 57 } 58 ans_ew = INF; ans_pw = 1; 59 dfs(1, 0, 0); 60 bool flag = false; 61 for(int i = 1; i <= n; ++i) { 62 if(!ans[i]) continue; 63 if(flag) printf(" "); 64 else flag = true; 65 printf("%d", i); 66 } 67 printf(" "); 68 } 69 }