hdu 2489 Minimal Ratio Tree【Dfs+kruskal】

Minimal Ratio Tree

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3928    Accepted Submission(s): 1207


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

1 3

1 2

 

 思路：

因为n比较小，所以我们可以采取直接暴力Dfs枚举所有可能选取的点的方案，然后累加点权值，然后将这些点相关的边加入数组中，贪心求最小生成树即可。

注意的点：我们在Dfs的时候，如果每一层都遍历n个点，是会超时的，所以我们枚举当前点是否加入子图中，这样就变成了每层枚举两个操作，防止超时的情况。

AC代码：

#include<string.h>
#include<algorithm>
#include<stdio.h>
using namespace std;
struct zuobiao
{
int x,y,w;
}a[121212];
int d[250];
int map[250][250];
int f[250];
int vis[250];
int output[250];
int n,m;
double minn;
int cmp(zuobiao a,zuobiao b)
{
return a.w<b.w;
}
int find(int a)
{
int r=a;
while(f[r]!=r)
r=f[r];
int i=a;
int j;
while(i!=r)
{
j=f[i];
f[i]=r;
i=j;
}
return r;
}
void merge(int a,int b)
{
int A,B;
A=find(a);
B=find(b);
if(A!=B)
f[B]=A;
}
void solve(int sumd)
{
int tot=0,sume=0;
for(int i=0;i<n;i++)
{
f[i]=i;
}
for(int i=0;i<n;i++)
{
if(vis[i]==1)
{
for(int j=0;j<n;j++)
{
if(vis[j]==0)continue;
if(map[i][j]==0)continue;
a[tot].x=i;
a[tot].y=j;
a[tot].w=map[i][j];
tot++;
}
}
}
sort(a,a+tot,cmp);
for(int i=0;i<tot;i++)
{
if(find(a[i].x)!=find(a[i].y))
{
merge(a[i].x,a[i].y);
sume+=a[i].w;
}
}
int tot2=0;
double out=sume*1.0/sumd;
if(out<minn||minn==-1)
{
minn=out;
for(int i=0;i<n;i++)
{
if(vis[i]==1)
output[tot2++]=i;
}
}
}
void Dfs(int cont,int now,int sumd)
{
if(now>n)return ;
if(cont==m)
{
solve(sumd);
return ;
}
vis[now]=1;
Dfs(cont+1,now+1,sumd+d[now]);
vis[now]=0;
Dfs(cont,now+1,sumd);
}
int main()
{
while(~scanf("%d%d",&n,&m))
{
if(n==0&&m==0)break;
for(int i=0;i<n;i++)
{
scanf("%d",&d[i]);
}
for(int i=0;i<n;i++)
{
for(int j=0;j<n;j++)
{
scanf("%d",&map[i][j]);
}
}
int ans[25];
memset(vis,0,sizeof(vis));
minn=-1;
Dfs(0,0,0);
for(int i=0;i<m;i++)
{
if(i==0)
printf("%d",output[i]+1);
else printf(" %d",output[i]+1);
}
printf("\n");
}
}
/*
3 2
30 10 20
0 3 1
3 0 1
1 1 0
//////1 3
3 2
30 30 10
0 3 1
3 0 1
1 1 0
//////1 3
3 2
30 80 10
0 3 1
3 0 1
1 1 0
//////2 3
*/