USACO sec2.1 Healthy Holsteins

状态压缩，先将生成的所有状态排序，然后枚举即可。

/*
PROG : holstein
LANG : C
*/
# include 
# include 

struct scoop{int a[30];};

int V, G, r[(1 << 17) + 5];
struct scoop g[17], v;

/***************************************************/
int bcnt(int x)
{
    int ret = 0;
    while (x)
    {
        ++ret;
        x &= x-1;
    }
    return ret;
}

int cmp(const void *xx, const void *yy)
{
    int x = *(int*)xx;
    int y = *(int*)yy;
    int t = bcnt(x) - bcnt(y);
    if (t == 0) return x>y ? 1:-1;
    return t;
}

void pre(void)
{
    int i;
    for (i = 0; i < (1<<(G+1)); ++i) r[i] = i;
    qsort(r, (1<<(G+1)), sizeof(r[1]), cmp);
}

int cal(int x)
{
    int i, j;
    struct scoop tmp = v;
    for (i = 0; i < G; ++i) if ((x>>i) & 0x1)
    for (j = 1; j <= V; ++j)
        tmp.a[j] -= g[i+1].a[j];
    for (j = 1; j <= V; ++j)
        if (tmp.a[j] > 0) return 0;
    return 1;
}

void solve(void)
{
    int i, k;
    pre();
    for (i = 0; i < (1<<(G+1)); ++i)
        if (cal(r[i])) break;
    printf("%d", bcnt(r[i]));
    for (k = 0; k < G; ++k) if ((r[i]>>k)&0x1)
        printf(" %d", k+1);
    putchar('\n');
}
/***************************************************/

void read(void)
{
    int i, j;
    scanf("%d", &V);
    for (i = 1; i <= V; ++i) scanf("%d", &v.a[i]);
    scanf("%d", &G);
    for (i = 1; i <= G; ++i)
    for (j = 1; j <= V; ++j)
        scanf("%d", &g[i].a[j]);
}

int main()
{
    freopen("holstein.in", "r", stdin);
    freopen("holstein.out", "w", stdout);

    read();
    solve();

    fclose(stdin);
    fclose(stdout);

    return 0;
}

DFS 毫无疑问会超时。

转载于:https://www.cnblogs.com/JMDWQ/archive/2012/08/21/2649482.html