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.
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.
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.
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】
【Sample Output】
【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。
#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue> using namespace std; bool flag[],outp[];
int n,m,node[];
int ma[][];
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[];
priority_queue<heaptype>heap;
heaptype aaa; memset(fla,,sizeof(fla));
fla[s]=true;
ans=;
for (i=;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=;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=;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=;i<=n;i++) outp[i]=fla[i];
}
} void dfs(int now,int last,int tot)
{
if (now==m)
{
int i;
for (i=;i<=n;i++)
if (flag[i]) break;
prim(i,tot);
}
else
{
now++;
for (int i=last+;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==&&m==))
{
for (int i=;i<=n;i++) scanf("%d",&node[i]);
for (int i=;i<=n;i++)
for (int j=;j<=n;j++) scanf("%d",&ma[i][j]); mi=;
memset(flag,,sizeof(flag));
for (int i=;i<=n-m+;i++)
{
flag[i]=true;
dfs(,i,node[i]);
flag[i]=false;
} int i;
for (i=;i<=n;i++)
if (outp[i])
{
printf("%d",i);
break;
}
for (int j=i+;j<=n;j++)
if (outp[j]) printf(" %d",j);
printf("\n");
scanf("%d%d",&n,&m);
} return ;
}