hdu2489 Minimal Ratio Tree dfs枚举组合情况+最小生成树

#include <stdio.h>
#include <set>
#include <string.h>
#include <algorithm>
using namespace std;

const int maxn = 30;
const int inf = 999999999;
double minans;
int vis[maxn];//记录选中的点
int mp[maxn][maxn], ans[maxn][maxn];
int min_road[maxn];

int Prim(int n)
{
    int i, j, min_i, minm, sum = 0;
    int dis[maxn];
    int vis[maxn];
    for (i = 1; i <= n; i++)
        dis[i] = ans[i][1];
    memset(vis, false, sizeof vis);
    vis[1] = true;
    for (i = 1; i<n; i++)
    {
        minm = inf, min_i = i;
        for (j = 1; j <= n; j++)
        {
            if (vis[j] == false && dis[j]<minm)
            {
                minm = dis[j];
                min_i = j;
            }
        }
        if (minm == inf)
            break;
        sum += minm;
        vis[min_i] = true;
        for (j = 1; j <= n; j++)
        {
            if (vis[j] == 0 && dis[j]>ans[min_i][j])
                dis[j] = ans[min_i][j];
        }
    }
    return sum;
}

int ok(int n)
{
    memset(vis, 0, sizeof vis);
    int cnt = 1;
    int m = n, cont = 1;
    while (m)
    {
        if (m % 2)
            vis[cont++] = cnt;
        cnt++;
        m /= 2;
    }
    return cont;
}

int main()
{
    int d[maxn], maxm;
    int n, i, j, k, cnt, m;
    while (scanf("%d%d", &n, &m) != EOF)
    {
        if (n == 0 && m == 0)
            break;
        maxm = 1;
        for (i = 1; i <= n; i++)
            maxm *= 2;
        for (i = 1; i <= n; i++)
            scanf("%d", &d[i]);
        for (i = 1; i <= n; i++)
        {
            for (j = 1; j <= n; j++)
            {
                scanf("%d", &mp[i][j]);
            }
        }
        minans = 9999999999.0;
        int sum_point;
        for (i = 0; i < maxm; i++)
        {
            if (ok(i) == m + 1)
            {
                sum_point = 0;
                for (j = 1; j <= m; j++)
                {
                    sum_point += d[vis[j]];//选的点
                    for (k = j + 1; k <= m; k++)
                    {
                        ans[j][k] = ans[k][j] = mp[vis[j]][vis[k]];
                    }
                }
                int sum = Prim(m);
                if ((sum*1.0 / sum_point) < minans)
                {
                    minans = sum*1.0 / sum_point;
                    for (i = 1; i <= m; i++)
                        min_road[i] = vis[i];
                }
            }
        }
        for (i = 1; i <= m - 1; i++)
            printf("%d ", min_road[i]);
        printf("%d
", min_road[i]);
    }
    return 0;
}