Minimal Ratio Tree

Minimal Ratio Tree

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 32768/32768K (Java/Other)
Total Submission(s) : 28 Accepted Submission(s) : 15
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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 110int 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;}