HDU - 2489 + 最小生成树+暴力组合选数

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.

InputInput 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.

OutputFor 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

Sample Output

1 3

1 2

思路: 要求从n中选m个组成的一颗最小生成树（也就是边权最小sum）,并且要在此条件下点权最小，也就是sum/ans最小，那么可以暴力枚举出从n个点在选m个，然后取sum/ans最小的那个；

思路可以参见:https://blog.csdn.net/xingyeyongheng/article/details/9373271

组合选数参见:https://blog.csdn.net/qq_34649947/article/details/80408580

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
using namespace std;
typedef long long ll;
const int INF = 0x3f3f3f3f;
const int maxn = 100 + 10;
const double eps = 1e-8;
int cost[maxn][maxn];//表示边权值，最小生成树算法用到
int mincost[maxn];
bool used[maxn];
int V;//顶点数 == m
int n, m;
int weight[maxn];//每个点的权值
int record[maxn];//每次选取的m个点的集合
int result[maxn];//最后的结果集
int prim() {
    //注意！！！这里的i-v的关系，prim的集合应该是record里面的，不是i下标
    memset(used, false, sizeof(used));
    for (int i = 1; i <= m; i++) {
        int v = record[i];
        mincost[v] = INF;
        used[v] = false;
    }
    mincost[record[1]] = 0;
    int res = 0;
    while (true) {
        int u = -1;
        //从不属于最小生成树集合X的顶点中选取从当前已经生成的树X到其权值最小的顶点
        for (int i = 1; i <= m; i++) {
            if (!used[record[i]] && (u == -1 || mincost[record[i]] < mincost[u])) u = record[i];
        }
        if (u == -1) break;
        used[u] = true;//把顶点加入到X
        res += mincost[u];//把边的长度加入到结果里

        //更新当前u的领边的情况
        for (int i = 1; i <= m; i++) {
            int v = record[i];
            mincost[v] = min(mincost[v], cost[u][v]);
        }
    }
    return res;
}
double minratio = INF*1.0;//
void dfs(int k, int dp) {
    if (dp == m) {
        int ans = 0;
//        for (int i = 1; i <= m; i++) {
//            cout << record[i] << " ";
//        }
        for (int i = 1; i <= m; i++) ans += weight[record[i]];
        int a = prim();
//        cout << a << " " << ans << endl;
        double sum = a*1.0/ans;
//        cout << sum << endl;
        if (sum < minratio) {
            minratio = sum;
            for (int i = 1; i <= m; i++) result[i] = record[i];
        }
        return;
    }
    //若剩余的点的个数(n-k)加上目前选取的个数num小于m的话，说明即使接下来n-k个点都选取，也选不足m个点，直接return
    if (n - k + dp < m) return;
    for (int i = k + 1; i <= n; i++) {
        record[dp + 1] = i;
        dfs(i, dp + 1);
    }
}
int main() {
    while (~scanf("%d %d", &n, &m)) {
        if (n == 0 && m == 0) break;
//        V = m;//最小生成树的点的个数
        //输入每个点的权值
        //!!!!!!!不要忘记清空初始化
        memset(result, 0, sizeof(result));
        memset(weight, 0, sizeof(weight));
        memset(cost, INF, sizeof(cost));
        memset(record, 0, sizeof(record));
        minratio = INF*1.0;
        for (int i = 1; i <= n; i++) {
            scanf("%d", &weight[i]);
        }
        //输入边权
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= n; j++) {
                scanf("%d", &cost[i][j]);
            }
        }
        //n中选取m个点
        for (int i = 1; i <= n; i++) {
            record[1] = i;
            dfs(i, 1);
        }
        for (int i = 1; i <= m; i++) {
            if (i == 1) printf("%d", result[i]);
            else printf(" %d", result[i]);
        }
        printf("\n");
    }
    return 0;
}

HDU - 2489 + 最小生成树+暴力组合选数

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