标签：hdu2489
Minimal Ratio Tree
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2382 Accepted Submission(s): 709
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.
Output
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 .
Sample Input
3 2
30 20 10
0 6 2
6 0 3
2 3 0
2 2
1 1
0 2
2 0
0 0
Sample Output
Source
⊙﹏⊙‖∣在判断两浮点数大小时应该这样比较：a-b<-(1e-8)；我因为不知道这个WA了6次。
题意：求一个略微变形的“最小生成树”，其值为边权和除以点权和。
题解：用深搜在n个点里选出m个点，再求这m个点的“最小生成树”即可。
#include <stdio.h>
#include <string.h>
#include <limits.h>
#define maxn 16
int map[maxn][maxn], node[maxn];
int n, m, store[maxn], vis[maxn];
double ans;
bool visted[maxn]; //hash to vis array
double prim()
{
int i, j, u, count = 0, tmp, vnv = 0, vne = 0;
for(i = 1; i <= m; ++i) vnv += node[vis[i]];
memset(visted, 0, sizeof(visted));
visted[1] = 1;
while(count < m - 1){
for(i = 1, tmp = INT_MAX; i <= m; ++i){
if(!visted[i]) continue;
for(j = 1; j <= m; ++j){
if(!visted[j] && map[vis[i]][vis[j]] < tmp){
tmp = map[vis[i]][vis[j]]; u = j;
}
}
}
if(tmp != INT_MAX){
visted[u] = 1;
vne += tmp; ++count;
}
}
return vne * 1.0 / vnv;
}
void DFS(int k, int id)
{
if(id > m){
double tmp = prim();
if(tmp - ans < -(1e-8)){
ans = tmp; memcpy(store, vis, sizeof(vis));
}
return;
}
for(int i = k; i <= n; ++i){
vis[id] = i;
DFS(i + 1, id + 1);
}
}
int main()
{
int i, j;
while(scanf("%d%d", &n, &m), n || m){
for(i = 1; i <= n; ++i) scanf("%d", &node[i]);
for(i = 1; i <= n; ++i)
for(j = 1; j <= n; ++j)
scanf("%d", &map[i][j]);
ans = INT_MAX;
DFS(1, 1);
for(i = 1; i <= m; ++i)
if(i != m) printf("%d ", store[i]);
else printf("%d\n", store[i]);
}
return 0;
}
HDU2489 Minimal Ratio Tree 【DFS】+【最小生成树Prim】,布布扣,bubuko.com
HDU2489 Minimal Ratio Tree 【DFS】+【最小生成树Prim】
标签：hdu2489
原文地址：http://blog.csdn.net/chang_mu/article/details/38471659
