hdu 2489（枚举 + 最小生成树）

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

思路：由于N, M的范围比较少，直接枚举所有的可能情况，然后求MST判断即可。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <vector>
using namespace std;

const int MAX_N = 17;
const double eps = 1e-10;
struct Edge {
	int u, v, w;
	Edge() {}
	Edge(int _u, int _v, int _w) : u(_u), v(_v), w(_w) {}
} edge[MAX_N * MAX_N];

int cmp(const Edge &e1, const Edge &e2) 
{
	return e1.w < e2.w;
}

int N, M, cnt, weight[MAX_N], g[MAX_N][MAX_N];
int parent[MAX_N];
void Init()
{
	for (int i = 0; i < N; ++i) parent[i] = i;
}

int find(int x)
{
	if (x == parent[x]) return x;

	return parent[x] = find(parent[x]);
}

void Union(int u, int v)
{
	int r1 = find(u), r2 = find(v);
	if (r1 == r2) return;

	parent[r1] = r2;
}

double res;
vector<int > tmp, ans;
bool vis[MAX_N];
void gao(int state)
{
	tmp.clear();
	memset(vis, false, sizeof(vis));
	int a = 0, b = 0;
	for (int i = 0; i < N; ++i) {
		if ((1 << i) & state) tmp.push_back(i), b += weight[i], vis[i] = true;
	}

	Init();
	for (int i = 0; i < cnt; ++i) {
		Edge e = edge[i];
		if (vis[e.u] && vis[e.v]) {
			if (find(e.u) != find(e.v)) {
				Union(e.u, e.v);
				a += e.w;
			}
		}
	}

	if (res < eps) {
		res = a * 1.0 / b;
		ans = tmp;
	} else if (a * 1.0 / b + eps < res) {
		res = a * 1.0 / b;
		ans = tmp;
	} else if (fabs(a * 1.0 / b - res) < eps) {
		bool tag = false;
		for (int i = 0; i < (int)ans.size(); ++i) {
			if (ans[i] > tmp[i]) { tag = true; break; }
		}
		
		if (tag) ans = tmp;
	}
	
}

int main()
{
	while (~scanf("%d %d", &N, &M)) {
		if (N == 0 && M == 0) break;

		for (int i = 0; i < N; ++i) scanf("%d", &weight[i]);
		for (int i = 0; i < N; ++i) {
			for (int j = 0; j < N; ++j) scanf("%d", &g[i][j]);
		}
		
		cnt = 0;
		for (int i = 0; i < N; ++i) {
			for (int j = i + 1; j < N; ++j) edge[cnt++] = Edge(i, j, g[i][j]);
		}

		sort(edge, edge + cnt, cmp);
		res = 0.0;
		ans.clear();
		for (int s = 0; s < (1 << N); ++s) {
			int x = s, n = 0;
			while (x) {
				++n;
				x &= (x - 1);
			}
			
			if (n == M) {
				gao(s);
			}
		}

		for (int i = 0; i < (int)ans.size(); ++i) {
			printf(i == 0 ? "%d" : " %d", ans[i] + 1);
		}
		puts("");
		
	}
	return 0;
}