Minimal Ratio Tree
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 32768/32768K (Java/Other) Total Submission(s) : 28 Accepted Submission(s) : 15 Font: Times New Roman | Verdana | Georgia Font Size: ← → Problem 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 . Sample Input 3 2 30 20 10 0 6 2 6 0 3 2 3 0 2 2 1 1 0 2 2 0 0 0 Sample Output 1 3 1 2 Source 2008 Asia Regional Beijing
#include<iostream> #include<cstdio> #include<iomanip> #include<algorithm> #include<cstring> #include<cmath> using namespace std; #define INF 999999 #define maxsize 110 int n, m;bool vist[maxsize]; int map[maxsize][maxsize]; int Nodep[maxsize]; int dis[maxsize]; int ans[maxsize]; bool chose[maxsize]; double minn; double Prim(int f) { int i, j, x; memset(vist, 0, sizeof(vist)); for (i = 0;i < 20;i++) dis[i] = 10000000; dis[f] = 0; int minx, min, ansx = 0;; for (i = 1;i <= m;i++) { min = 100000000; for (j = 1;j <= n;j++) if (chose[j] && !vist[j] && dis[j] < min) { min = dis[j]; minx = j; } vist[minx] = 1; ansx += dis[minx]; dis[minx] = 0; for (j = 1;j <= n;j++) if (chose[j] && !vist[j] && map[minx][j] < dis[j]) dis[j] = map[minx][j]; } return ansx * 1.0; } void Dfs(int p, int c) { if (c == m) { int first = 0;; int countn = 0; double tem, tem1, tem2 = 0; for (int i = 1;i <= n;i++) if (chose[i]) { if (!first) { first = i; tem1 = Prim(first); } tem2 += Nodep[i] * 1.0; } tem = tem1 / tem2; if (tem < minn) { minn = tem; for (int i = 1;i <= n;i++) if (chose[i]) { ans[countn] = i; countn++; } } return; } chose[p] = 1; Dfs(p + 1, c + 1); chose[p] = 0; if (n - p + c >= m) Dfs(p + 1, c); } int main() { while (cin >> n >> m&&n&&m) { int t; //memset(Nodep, 0, sizeof(dis)); memset(chose, 0, sizeof(chose)); minn = INF; for (int i = 1;i <= n;i++) { cin >> Nodep[i]; } //memset(map, 0, sizeof(map)); for (int i = 1;i <= n;i++) { for (int j = 1;j <= n;j++) { cin >> t; map[i][j] = t; } } Dfs(1, 0); for (int i = 0;i < m - 1;i++) { cout << ans[i] << " "; } cout << ans[m-1] << endl; } return 0; }