杭电oj HDOJ 2489 Minimal Ratio Tree（DFS 最小生成树）

杭电oj HDOJ 2489 Minimal Ratio Tree

题目来源：http://acm.hdu.edu.cn/showproblem.php?pid=2489

Problem Description

For a tree, which nodes and edges are all weighted, the ratio of it is calculated according to the following equation.

Given a complete graph of n nodes with all nodes and edges weighted, your task is to find a tree, which is a sub-graph of the original graph, with m nodes and whose ratio is the smallest among all the trees of m nodes in the graph.

Input

Input contains multiple test cases. The first line of each test case contains two integers n (2<=n<=15) and m (2<=m<=n), which stands for the number of nodes in the graph and the number of nodes in the minimal ratio tree. Two zeros end the input. The next line contains n numbers which stand for the weight of each node. The following n lines contain a diagonally symmetrical n×n connectivity matrix with each element shows the weight of the edge connecting one node with another. Of course, the diagonal will be all 0, since there is no edge connecting a node with itself.

All the weights of both nodes and edges (except for the ones on the diagonal of the matrix) are integers and in the range of [1, 100].

The figure below illustrates the first test case in sample input. Node 1 and Node 3 form the minimal ratio tree.

Ouput

For each test case output one line contains a sequence of the m nodes which constructs the minimal ratio tree. Nodes should be arranged in ascending order. If there are several such sequences, pick the one which has the smallest node number; if there’s a tie, look at the second smallest node number, etc. Please note that the nodes are numbered from 1 .

题目大意

一棵树的比率定义为“若干条边的权重之和”除以“这些条边之间点的权重之和”。现给出一个$n\times n$的图，要求求出一个生成树使其点有m个并且其比率最小。从大到小输出这棵树的结点编号。

解题思路

由于n的范围是2到15所以枚举出所有m个结点的树也不会超时。因此利用DFS枚举出所有的情况，并对每一种情况使用Prim算法求出其最小生成树，保存其最小比率树的结点，最后输出出来。

C++代码

#include 
#include 
#include 
#include 
#define inf 0x3f3f3f
using namespace std;

int graph[16][16];	
int nodeWeight[16];				// 结点权重数组
int dist[16];					// 结点到已确定集合的距离
bool mark[16];
int n, m;
vector<int> v, min_v;
vector<vector<int>> V;
void init(vector<int>);			// 初始化距离数组函数
void Prim();					// Prim算法，参数为循环的次数
double radio(vector<int>);		// 计算比率
// 深度优先搜索
// 第一个参数表示目前查看的元素下标
// 第二个参数表示目前选中的元素数量
void DFS(int, int);

int main()
{
	int i, j;
	double min_radio, temp;
	while (cin >> n >> m) {
		if (!n) {
			break;
		}
		for (i = 1; i <= n; i++) {
			cin >> nodeWeight[i];
		}
		for (i = 1; i <= n; i++) {
			for (j = 1; j <= n; j++) {
				cin >> graph[i][j];
			}
		}
		// 使用深度深度优先搜索枚举出所有符合要求的情况
		DFS(1, 0);
		// 每一种情况计算一次最小生成树
		for (i = 0; i < V.size(); i++) {
			init(V[i]);
			Prim();
			for (j = 1; j <= n; j++) {
				temp = radio(V[i]);
				if (!i) {
					min_radio = temp;
					min_v = V[i];
				}
				else {
					if (temp < min_radio) {
						min_radio = temp;
						min_v = V[i];
					}
				}
			}
		}
		for (i = 0; i < m; i++) {
			if (i) {
				cout << " ";
			}
			cout << min_v[i];
		}
		cout << endl;
		V.clear();
	}
}

void init(vector<int> v) {
	// 先把所有距离数组清零
	memset(dist, 0, sizeof(dist));
	// 先把所有标记数组标记为已确定
	memset(mark, 1, sizeof(mark));
	// 假设第一个点为首先确定的点，所以从数组的第二个开始循环
	for (int i = 1; i < m; i++) {
		dist[v[i]] = graph[v[0]][v[i]];
		mark[v[i]] = false;
	}
}

void Prim() {
	int i, j, temp, pos;
	for (i = 1; i < m; i++) {
		temp = inf;
		// 找出当前距离数组中最小的一项
		for (j = 1; j <= n; j++) {
			if (!mark[j] && dist[j] < temp) {
				temp = dist[j];
				pos = j;
			}
		}
		mark[pos] = true;
		for (j = 1; j <= n; j++) {
			if (!mark[j]) {
				// 更新距离数组
				dist[j] = min(dist[j], graph[pos][j]);
			}
		}
	}
}

double radio(vector<int> v) {
	int i,	numer = 0, denom = 0;
	for (i = 0; i < m; i++) {
		numer += dist[v[i]];
		denom += nodeWeight[v[i]];
	}
	return double(numer) / denom;
}

void DFS(int num, int count)
{
	if (num > n) {
		// 如果满足结点的个数要求则把此情况保存起来
		if (count == m) {
			for (int i = 1; i <= n; i++) {
				if (mark[i]) {
					v.push_back(i);
				}
			}
			V.push_back(v);
			v.clear();
		}
	}
	else {
		mark[num] = true;
		DFS(num + 1, count + 1);
		mark[num] = false;
		DFS(num + 1, count);
	}
}

发现问题欢迎指出和纠正，谢谢！