HDU 2489 Minimal Ratio Tree 最小生成树+DFS

Minimal Ratio Tree

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Problem Description

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

1

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.

2

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

题意

给出一张n个点的图,图中的每一个结点以及每一条边都有其权值,要求从中选出m个点,找到m-1条边将其连接,使得边权值与点权值的比值达到最小。

分析

要使得比值最小,则点权值和尽可能地大同时边权值和尽可能地小。直接上考虑,边权值和尽可能小即对这m个点作最小生成树。

而题目给定的n不大,故可以用DFS搜出需要的m个点,然后对m个点进行最小生成树,中间注意判断和保存即可。

我用了一个dijkstra+优先队列的prim去找MST,这也是我第一次尝试使用STL的优先队列。

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#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>

using namespace std;

bool flag[16],outp[16];
int n,m,node[16];
int ma[16][16];
double mi;

typedef struct heaptyp
{
int num,key;
friend bool operator < (heaptyp a,heaptyp b)
{
return a.num>b.num;
}
} heaptype;

void prim(int s,int tot)
{
int i,j,now,ans;
bool fla[16];
priority_queue<heaptype>heap;
heaptype aaa;

memset(fla,0,sizeof(fla));
fla[s]=true;
ans=0;
for (i=1;i<=n;i++)
if (flag[i]&&ma[s][i])
{
heaptype temp;
temp.num=ma[s][i];
temp.key=i;
heap.push(temp);
aaa=heap.top();
}

for (j=1;j<m;j++)
{
heaptype h=heap.top();
heap.pop();
aaa=heap.top();
while (fla[h.key])
{
h=heap.top();
heap.pop();
}
now=h.key;
fla[now]=true;
ans+=h.num;
for (i=1;i<=n;i++)
if (flag[i]&&ma[now][i])
if (!fla[i])
{
heaptype temp;
temp.num=ma[now][i];
temp.key=i;
heap.push(temp);
aaa=heap.top();
}
}
double rat=(double)ans/tot;
if (mi-rat>0.0000001)
{
mi=rat;
for (int i=1;i<=n;i++) outp[i]=fla[i];
}
}

void dfs(int now,int last,int tot)
{
if (now==m)
{
int i;
for (i=1;i<=n;i++)
if (flag[i]) break;
prim(i,tot);
}
else
{
now++;
for (int i=last+1;i<=n-m+now;i++)
{
flag[i]=true;
dfs(now,i,tot+node[i]);
flag[i]=false;
}
}
}

int main()
{
scanf("%d%d",&n,&m);
while (!(n==0&&m==0))
{
for (int i=1;i<=n;i++) scanf("%d",&node[i]);
for (int i=1;i<=n;i++)
for (int j=1;j<=n;j++) scanf("%d",&ma[i][j]);

mi=2147483647;
memset(flag,0,sizeof(flag));
for (int i=1;i<=n-m+1;i++)
{
flag[i]=true;
dfs(1,i,node[i]);
flag[i]=false;
}

int i;
for (i=1;i<=n;i++)
if (outp[i])
{
printf("%d",i);
break;
}
for (int j=i+1;j<=n;j++)
if (outp[j]) printf(" %d",j);
printf("\n");
scanf("%d%d",&n,&m);
}

return 0;
}
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