{POJ}{3925}{Minimal Ratio Tree}{最小生成树}

题意：给定完全无向图，求其中m个子节点，要求Sum(edge)/Sum(node)最小。

思路：由于N很小，枚举所有可能的子节点可能情况，然后求MST，memset()在POJ G++里面需要cstring头文件。

#include <iostream>
#include <vector>
#include <map>
#include <cmath>
#include <memory>
#include <algorithm>
#include <cstdio>
#include <cstdlib>
using namespace std;

const int MAXN = 100;
const int INF = 1<<30;

#define CLR(x,y) memset(x,y,sizeof(x))
#define MIN(m,v) (m)<(v)?(m):(v)
#define MAX(m,v) (m)>(v)?(m):(v)
#define ABS(x) ((x)>0?(x):-(x))
#define rep(i,x,y) for(i=x;i<y;++i)

int n,m,k;

double ans = 0;

int select[MAXN];
int g[MAXN][MAXN];
int val[MAXN];
int visit[MAXN];
double dist[MAXN];
int ind[MAXN];

double Prim()
{
	int i,j,tmp,mark_i;
	int mark_min;
	int sum_node = 0;
	int sum_edge = 0;

	for(i = 0; i < n; ++i)
	{
		dist[i] = INF;
		visit[i] = 0;
	}

	for(i = 0; i < n; ++i)
		if(select[i]>0)
		{
			dist[i] = 0;
			break;
		}

	int cnt = 0;

	for(i = 0; i < n; ++i,++cnt)
	{
		if(cnt>=m) break;

		mark_min = INF;

		for(j  = 0; j < n; ++j)
		{
			if(select[j]>0 && !visit[j] && mark_min>dist[j])
			{
				mark_min = dist[j];
				mark_i = j;
			}
		}

		visit[mark_i] = 1;

		sum_edge += dist[mark_i];

		for( j = 0; j < n; ++j)
		{
			if(visit[j]==0 && select[j]>0 && dist[j] > g[mark_i][j])
				dist[j] = g[mark_i][j];
		}
	}

	for( i = 0; i < n; ++i)
		if(select[i] > 0 )
			sum_node += val[i];

	return double(sum_edge)/sum_node;

}

int DFS(int cur, int deep)
{
	double res = INF;

	select[cur] = 1;

	if(deep < m)
	{
		for(int i = cur+1; i < n; ++i)
		{
			DFS(i,deep+1);
		}
	}

	if(deep == m)
	{
		res = Prim();
		if(res < ans)
		{
			ans = res;
			for(int i = 0; i < n; ++i)
				ind[i] = select[i];

		}
	}

	select[cur] = 0;

	return 0;
}

int Solve()
{
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		if(n == 0 && m == 0) break;
		for(int i = 0 ; i < n; ++i)
			scanf("%d",&val[i]);
		for(int i = 0; i < n; ++i)
			for(int j = 0; j < n; ++j)
			{
				scanf("%d",&g[i][j]);
			}

		CLR(select,0);
		ans = INF;

		for(int i = 0 ; i < n; ++i)
			DFS(i,1);

		int tag = 0;
		for(int i = 0; i < n; ++i)
			if(ind[i] > 0)
				if(tag == 0)
				{
					printf("%d",i+1);
					tag = 1;
				}
				else
					printf(" %d",i+1);
		printf("\n");
	}
	return 0;
}

int main()
{
	Solve();

	return 0;
}