HDU-2489 Minimal Ratio Tree（最小生成树[Prim]）

Minimal Ratio Tree

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)

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

3 230 20 100 6 26 0 32 3 02 21 10 22 00 0

Sample Output

1 31 2

好多次都能想到正解，但是不敢去写，还是需要勇于尝试，不断地尝试才能有更多的理解。

数据范围很小，枚举每一种点的组合情况即可。

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const double EPS=0.000001;int ans,sume,sumv;int n,u,m,w[17],g[17][17];int ori,cnt,vis,dis[17],pre;double minr,tmp;void Prim() {    int i,j;    memset(dis,0x3f,sizeof(dis));    vis=ori,sume=sumv=0;    for(i=1;i<=n;++i)        if((vis&(1<<i))==0) {            dis[i]=0;            break;        }    for(j=1;j<=m;++j) {        u=0;        for(i=1;i<=n;++i)            if(!(vis&(1<<i))&&dis[u]>dis[i])                u=i;        vis|=1<<u;        sume+=dis[u];        sumv+=w[u];        for(i=1;i<=n;++i)            dis[i]=min(dis[i],g[u][i]);    }    if(minr-EPS>(tmp=double(sume)/sumv)) {        minr=tmp;        ans=ori;    }}void DFS(int i) {    if(cnt==pre) {        Prim();        return ;    }    if(i>n)        return ;    DFS(i+1);    ++cnt;    ori|=1<<i;    DFS(i+1);    --cnt;    ori&=~(1<<i);}int main() {    int i,j;    while(scanf("%d%d",&n,&m),n||m) {        for(i=1;i<=n;++i)            scanf("%d",&w[i]);        for(i=1;i<=n;++i)            for(j=1;j<=n;++j)                scanf("%d",&g[i][j]);        minr=999999999;        ori=cnt=0;        pre=n-m;        DFS(1);        for(i=1;i<=n;++i)            if(!(ans&(1<<i))) {                printf("%d",i++);                break;            }        for(;i<=n;++i)            if(!(ans&(1<<i)))                printf(" %d",i);        printf("\n");    }    return 0;}