POJ 3925 Minimal Ratio Tree 最小生成树

思路是枚举+最小生成树，用DFS枚举或者二进制枚举。

/*
ID: sdj22251
PROG: calfflac
LANG: C++
*/
#include <iostream>
#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <cmath>
#include <ctime>
#define MAX 100000000
#define LOCA
#define PI acos(-1.0)
using namespace std;
int d[16][16], node[16], v[16], dis[16], ans[16];
int n, m;
double mi;
bool used[16];
void search()
{
    int i, j, k, sum, weight;
    sum = 0;
    weight = 0;
    memset(used, false, sizeof(used));
    for(i = 0; i < m; i++)
        dis[v[i]] = MAX;
    for(i = 0; i < m; i++)
        dis[v[i]] = d[v[0]][v[i]];
    used[v[0]] = true;
    for(i = 1; i < m; i++)
    {
        int mini = MAX;
        int tmp = -1;
        for(j = 0; j < m; j++)
            if(!used[v[j]] && dis[v[j]] < mini)
            {
                mini = dis[v[j]];
                tmp = v[j];
            }
        used[tmp] = true;
        sum += mini;
        for(j = 0; j < m; j++)
        {
            if(!used[v[j]] && d[v[j]][tmp] < dis[v[j]])
                dis[v[j]] = d[v[j]][tmp];
        }
    }
    for(j = 0; j < m; j++)
        weight += node[v[j]];
    double T = (double)sum / (double)weight;
    if(T < mi)
    {
        for(j = 0; j < m; j++)
            ans[j] = v[j];
        mi = T;
    }
}
void dfs(int x, int cnt)
{
    v[cnt] = x;
    if(cnt == m - 1)
    {
        search();
        return;
    }
    for(int i = x + 1; i <= n; i++)
        dfs(i, cnt + 1);
}
int main()
{
#ifdef LOCAL
    freopen("ride.in","r",stdin);
    freopen("ride.out","w",stdout);
#endif
    int i, j;
    while(scanf("%d%d", &n, &m) != EOF)
    {
        if(n == 0 && m == 0) break;
        for(i = 1; i <= n; i++)
            scanf("%d", &node[i]);
        for(i = 1; i <= n; i++)
        {
            for(j = 1; j <= n; j++)
            {
                scanf("%d", &d[i][j]);
            }
        }
        mi = 100000000.0;
        for(i = 1; i <= n; i++)
        dfs(i, 0);
        for(i = 0; i < m - 1; i++)
            printf("%d ", ans[i]);
        printf("%d\n", ans[m - 1]);
    }
    return 0;
}