HDU 2489 Minimal Ratio Tree（暴力+最小生成树）（2008 Asia Regional Beijing）

Description

For a tree, which nodes and edges are all weighted, the ratio of it is calculated according to the following equation. Given a complete graph of n nodes with all nodes and edges weighted, your task is to find a tree, which is a sub-graph of the original graph, with m nodes and whose ratio is the smallest among all the trees of m nodes in the graph. 

 

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. 

 

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）：

 #include <cstdio>
 #include <cstring>
 #include <iostream>
 using namespace std;

 const int MAXN = ;
 const int INF = 0x3fff3fff;

 int mat[MAXN][MAXN];
 int weight[MAXN];
 int n, m;
 bool use[MAXN], vis[MAXN];
 int dis[MAXN];

 int prim(int st) {
     memset(vis, , sizeof(vis));
     vis[st] = true;
     for(int i = ; i <= n; ++i) dis[i] = mat[st][i];
     int ret = ;
     for(int cnt = ; cnt < m; ++cnt) {
         int u, min_dis = INF;
         for(int i = ; i <= n; ++i)
             if(use[i] && !vis[i] && dis[i] < min_dis) u = i, min_dis = dis[i];
         ret += min_dis;
         vis[u] = true;
         for(int i = ; i <= n; ++i)
             if(use[i] && !vis[i] && dis[i] > mat[u][i]) dis[i] = mat[u][i];
     }
     return ret;
 }

 bool ans[MAXN];
 int ans_pw, ans_ew;

 void dfs(int dep, int cnt, int sum_pw) {
     if(cnt == m) {
         int sum_ew = prim(dep - );
         if(ans_ew == INF || ans_ew * sum_pw > ans_pw * sum_ew) {//ans_ew/ans_pw > sum_ew/sum_pw
             for(int i = ; i <= n; ++i) ans[i] = use[i];
             ans_ew = sum_ew; ans_pw = sum_pw;
         }
         return ;
     }
     if(dep == n + ) return ;
     use[dep] = true;
     dfs(dep + , cnt + , sum_pw + weight[dep]);
     use[dep] = false;
     dfs(dep + , cnt, sum_pw);
 }

 int main() {
     while(scanf("%d%d", &n, &m) != EOF) {
         if(n ==  && m == ) break;
         for(int i = ; i <= n; ++i) scanf("%d", &weight[i]);
         for(int i = ; i <= n; ++i) {
             for(int j = ; j <= n; ++j) scanf("%d", &mat[i][j]);
         }
         ans_ew = INF; ans_pw = ;
         dfs(, , );
         bool flag = false;
         for(int i = ; i <= n; ++i) {
             if(!ans[i]) continue;
             if(flag) printf(" ");
             else flag = true;
             printf("%d", i);
         }
         printf("\n");
     }
 }