hdu - 2489 - Minimal Ratio Tree（枚举 + MST）

题意：给出一个图 n x n (2<=n<=15)的图，每个点，每条边都有权值，求其中的 m (2<=m<=n)个点，使得这m个点生成的树的边点权比例最小。

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2489

——>>数量小，于是，可以枚举取 m 个点的所有情况，对每种情况最一次MST，更新最小值。。

时间复杂度：O(n ^ n * log(n) * 2 ^ n)

#include <cstdio>
#include <cstring>
#include <queue>
#include <cmath>

using std::priority_queue;

const int MAXN = 15;
const int INF = 0x3f3f3f3f;
const double EPS = 1e-8;

struct EDGE
{
int from;
int to;
int w;

EDGE() {}
EDGE(int from, int to, int w) : from(from), to(to), w(w) {}

bool operator < (const EDGE& e) const
{
return w > e.w;
}
};

int n, m;
int sume, sumn;
int node[MAXN];
int G[MAXN][MAXN];
int toUse[MAXN], ucnt;
int fa[MAXN];

void Init()
{
sume = 0;
sumn = 0;
}

int Find(int x)
{
return x == fa[x] ? x : (fa[x] = Find(fa[x]));
}

void Union(int x, int y)
{
int xroot = Find(x);
int yroot = Find(y);

if (xroot != yroot)
{
fa[yroot] = xroot;
sume += G[x][y];
}
}

void Read()
{
for (int i = 0; i < n; ++i)
{
scanf("%d", node + i);
}
for (int i = 0; i < n; ++i)
{
for (int j = 0; j < n; ++j)
{
scanf("%d", &G[i][j]);
}
}
}

int Ones(int x)
{
int ret = 0;

while (x)
{
if (x & 1)
{
++ret;
}
x >>= 1;
}

return ret;
}

void GetUseNode(int S)
{
ucnt = 0;
for (int j = 0; (1 << j) <= S; ++j)
{
if ((1 << j) & S)
{
toUse[ucnt++] = j;
sumn += node[j];
}
}
}

void InitUnion()
{
for (int j = 0; j < n; ++j)
{
fa[j] = j;
}
}

void Kruscal()
{
priority_queue<EDGE> pq;
for (int j = 0; j < ucnt; ++j)
{
for (int k = j + 1; k < ucnt; ++k)
{
pq.push(EDGE(toUse[j], toUse[k], G[toUse[j]][toUse[k]]));
}
}
while (!pq.empty())
{
EDGE e = pq.top();
pq.pop();
Union(e.from, e.to);
}
}

void Output(int ret)
{
bool fst = true;
for (int i = 0; (1 << i) <= ret; ++i)
{
if ((1 << i) & ret)
{
if (fst)
{
fst = !fst;
}
else
{
putchar(' ');
}
printf("%d", i + 1);
}
}
puts("");
}

int Dcmp(double x)
{
if (fabs(x) < EPS) return 0;
return x > 0 ? 1 : -1;
}

void Solve()
{
int ret = 0;
double Min = INF;

for (int i = 3; i < (1 << n); ++i)
{
if (Ones(i) != m) continue;
Init();
GetUseNode(i);
InitUnion();
Kruscal();

double r = (double)sume / sumn;
if (Dcmp(r - Min) < 0)
{
Min = r;
ret = i;
}
}

Output(ret);
}

int main()
{
while (scanf("%d%d", &n, &m) == 2)
{
if (!n && !m) break;
Read();
Solve();
}
return 0;
}