HDU 2489 Minimal Ratio Tree

HDU_2489

    由于N很小，可以枚举保留的顶点的情况并依次计算最小生成树，然后更新一下记录的结果即可。

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#define MAXD 20
#define MAXM 400
int N, M, u[MAXM], v[MAXM], w[MAXM], vis[MAXD], r[MAXM], node[MAXD], p[MAXD], e, wa[MAXD], wb[MAXD], *a, *b, ew, nw;
int cmp(const void *_p, const void *_q)
{
    int *p = (int *)_p, *q = (int *)_q;
    return w[*p] - w[*q];
}
int find(int x)
{
    return p[x] == x ? x : (p[x] = find(p[x]));
}
void init()
{
    int i, j, k;
    for(i = 0; i < N; i ++)
        scanf("%d", &node[i]);
    e = 0;
    for(i = 0; i < N; i ++)
        for(j = 0; j < N; j ++)
        {
            u[e] = i, v[e] = j;
            scanf("%d", &k);
            if(i != j)
                w[e ++] = k;
        }
}
void update()
{
    int i, *c;
    for(i = 0; i < M; i ++)
    {
        if(a[i] < b[i])
            return ;
        else if(a[i] > b[i])
        {
            c = a, a = b, b = c;
            return ;
        }
    }
}
void getans()
{
    int i, j, k, x, y, t1, t2, *c;
    t1 = t2 = 0;
    for(i = 0; i < N; i ++)
        p[i] = i;
    for(i = 0; i < e; i ++)
    {
        k = r[i];
        if(!vis[u[k]] && !vis[v[k]])
        {
            x = find(u[k]), y = find(v[k]);
            if(x != y)
            {
                t1 += w[k];
                p[x] = y;
            }
        }
    }
    for(i = 0; i < M; i ++)
        t2 += node[b[i]];
    if(ew < 0 || t1 * nw < t2 * ew)
    {
        ew = t1, nw = t2;
        c = a, a = b, b = c;
    }
    else if(t1 * nw == t2 * ew)
        update();
}
void solve()
{
    int i, j, k, t;
    a = wa, b = wb, ew = -1;
    for(i = 0; i < e; i ++)
        r[i] = i;
    qsort(r, e, sizeof(r[0]), cmp);
    for(k = 0; k < (1 << N); k ++)
    {
        t = 0;
        for(i = 0; i < N; i ++)
        {
            if((1 << i) & k)
            {
                vis[i] = 0;
                b[t] = i;
                ++ t;
            }
            else
                vis[i] = 1;
        }
        if(t == M)
            getans();
    }
    printf("%d", a[0] + 1);
    for(i = 1; i < M; i ++)
        printf(" %d", a[i] + 1);
    printf("\n");
}
int main()
{
    for(;;)
    {
        scanf("%d%d", &N, &M);
        if(!N && !M)
            break;
        init();
        solve();
    }
    return 0;
}