hdu 2489 Minimal Ratio Tree(dfs枚举 + 最小生成树)~~~

题目：

        链接：点击打开链接

题意：

        输入n个点,要求选m个点满足连接m个点的m-1条边权值和sum与点的权值和ans使得sum/ans最小,并输出所选的m个点,如果有多种情况就选第一个点最小的,如果第一个点也相同就选第二个点最小的........

思路：

        求一个图中的一颗子树，使得Sum(edge weight)/Sum(point weight)最小~

        数据量小，暴力枚举~~~~~dfs暴力枚举C(M,N)种情况。

        枚举出这M个点之后，Sum(point weight)固定，进行prim或者Kruskal算法使Sum(edge weight)最小。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define MAX 10000000
const int N = 20;

int n,m;
int w[N],ans[N],node[N];
int map[N][N],f[N][N];
int vis[N];
int low[N];
double res;

int prim()
{
    int pos,minn;
    int result = 0;
    memset(vis,0,sizeof(vis));
    pos = 1;
    vis[pos] = 1;
    for(int i=1; i<=m; i++)
    {
        if(i != pos)
            low[i] = map[pos][i];
    }
    for(int i=1; i<m; i++)
    {
        minn = MAX;
        for(int j=1; j<=m; j++)
        {
            if(!vis[j] && minn > low[j])
            {
                minn = low[j];
                pos = j;
            }
        }
        result += minn;
        vis[pos] = 1;
        for(int j=1; j<=m; j++)
        {
            if(!vis[j] && map[pos][j]<low[j])
                low[j] = map[pos][j];
        }
    }
    return result;
}

void dfs(int u,int cnt)
{
    node[cnt] = u;
    if(cnt >= m)
    {
        for(int i=1; i<=m; i++)
        {
            for(int j=1; j<=m; j++)
                map[i][j] = f[node[i]][node[j]];
        }
        double SumEdge = 1.0 * prim();
        double SumPoint = 0;
        for(int i=1; i<=m; i++)
        {
            SumPoint += w[node[i]];
        }
        double ratio1 = SumEdge/SumPoint;
        if(ratio1 - res < -1e-8)
        {
            res = ratio1;
            for(int i=1; i<=m; i++)
                ans[i] = node[i];
        }
        return ;
    }
    for(int i=u+1; i<=n; i++)
        dfs(i,cnt+1);
}

int main()
{
    //freopen("input.txt","r",stdin);
    while(scanf("%d%d",&n,&m) != EOF && (n || m))
    {
        for(int i=1; i<=n; i++)
            scanf("%d",&w[i]);
        for(int i=1; i<=n; i++)
        {
            for(int j=1; j<=n; j++)
                scanf("%d",&f[i][j]);
        }
        res = MAX;
        for(int i=1; i<=n; i++)
        {
            dfs(i,1);
        }
        for(int i=1; i<m; i++)
            printf("%d ",ans[i]);
        printf("%d\n",ans[m]);
    }
    return 0;
}

------------------------------------------------------------------

战斗，从不退缩；奋斗，永不停歇~~~~~~~~~~~~

hdu 2489 Minimal Ratio Tree(dfs枚举 + 最小生成树)~~~,布布扣,bubuko.com

hdu 2489 Minimal Ratio Tree(dfs枚举 + 最小生成树)~~~

原文：http://blog.csdn.net/u013147615/article/details/29899515