HDU 2489 Minimal Ratio Tree（图论-最小生成树）

Minimal Ratio Tree

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2367    Accepted Submission(s): 707

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

3 230 20 100 6 26 0 32 3 02 21 10 22 00 0

 

Sample Output

1 31 2

 

Source

2008 Asia Regional Beijing

 

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gaojie

题目大意：

给你一张图n个点，每个点有权值，问你从中选出m个点，使得最小，输出方案。

解题思路：

用二进制取与不取来枚举选出m个点的方案，分母就确定了，分子越小越好，所以通过最小生成树算出。

参考代码：

#include <iostream>#include <cstring>using namespace std;const int MAXN = 20;const int INF = 0x3f3f3f3f;int graph[MAXN][MAXN], weight[MAXN], dis[MAXN], pre[MAXN], n, m, ans;long long ansEdge, ansNode;bool valid[MAXN], v[MAXN];int prim() {    memset(v, false, sizeof(v));    int u;    for (int i = 1; i <= n; i++) {        if (valid[i]) {            u = i;            break;        }    }    v[u] = true;    int edgeSum = 0;    for (int i = 1; i <= n; i++) {        if (valid[i]) {            dis[i] = graph[u][i];            pre[i] = u;        }    }    for (int i = 1; i < m; i++) {        u = -1;        for (int j = 1; j <= n; j++) {            if (valid[j] && !v[j]) {                if (u == -1 || dis[u] > dis[j])                    u = j;            }        }        edgeSum += graph[pre[u]][u];        v[u] = true;        for (int j = 1; j <= n; j++) {            if (valid[j] && !v[j]) {                if (dis[j] > graph[u][j]) {                    dis[j] = graph[u][j];                    pre[j] = u;                }            }        }    }    return edgeSum;}void init() {    ansEdge = INF;    ansNode = 0;}void input() {    for (int i = 1; i <= n; i++) {        cin >> weight[i];    }    for (int i = 1; i <= n; i++) {        for (int j = 1; j <= n; j++) {            cin >> graph[i][j];            if (graph[i][j] == 0) {                graph[i][j] = INF;            }        }    }}void work() {    for (int i = (1 << m) - 1; i < (1 << n); i++) {        int cnt = 0;        long long edge_cur = 0, node_cur = 0;        memset(valid, false, sizeof(valid));        for (int j = 1; j <= n; j++) {            if (i & (1 << (j-1))) {                cnt++;                valid[j] = true;                node_cur += weight[j];            }        }        if (cnt == m) {            edge_cur = prim();            if (edge_cur * ansNode < ansEdge * node_cur) {                ansEdge = edge_cur;                ansNode = node_cur;                ans = i;            }        }    }}void output() {    int cnt = 0;    for (int i = 1; i <= n; i++) {        if (ans & (1 << (i-1))) {            if (cnt++) cout << " ";            cout << i;        }    }    cout << endl;}int main() {    ios::sync_with_stdio(false);    while (cin >> n >> m && (n || m)) {        init();        input();        work();        output();    }    return 0;}