hdu 2489 Minimal Ratio Tree

https://vjudge.net/problem/HDU-2489

题意：求一个完全图的最优比率生成树，点的个数由题给出。最优比率生成树是边的权值之和与点的权值之和的比值最小的生成树。

思路：一开始用dfs枚举搜索每一种情况，t了，枚举的情况太多。之后看了题解，用的是状态压缩的方法枚举点的选择，这样的情况要少得多。至于其他的细节倒是很好想，由于比值最小，所以我们要求的是一棵最小生成树，再比上点的权值之和，最后选择出最小的就行了。

#include <stdio.h>
#include <string.h>
#include <map>
#include <algorithm>
using namespace std;

int n,m;

int par[20];
int a[20][20];
int b[20];

struct node
{
    int x,y,d;
} g[1000];

void init(int k)
{
    for (int i = 0;i <= k;i++)
        par[i] = i;
}

int fin(int x)
{
    if (x == par[x]) return x;
    else return par[x] = fin(par[x]);
}

void unit(int x,int y)
{
    x = fin(x);
    y = fin(y);

    if (x != y) par[x] = y;
}

bool cmp(node aa,node bb)
{
    return aa.d < bb.d;
}

double solve(int k)
{
    init(n);

    int p[20],cnt = 0;
    int num = 0;

    for (int i = 0;i < n;i++)
    {
        if (k & (1 << i)) p[cnt++] = i;
    }

    for (int i = 0;i < cnt;i++)
    {
        for (int j = i + 1;j < cnt;j++)
        {
            g[num].x = p[i];
            g[num].y = p[j];
            g[num].d = a[p[i]][p[j]];
            num++;
        }
    }

    int sum = 0,sm = 0;

    for (int i = 0;i < cnt;i++)
    {
        int t = p[i];
        sum += b[t];
    }

    sort(g,g+num,cmp);

    for (int i = 0;i < num;i++)
    {
        int x = g[i].x,y = g[i].y;

        //printf("%d %d88
",x,y);

        if (fin(x) == fin(y)) continue;

        unit(x,y);

        sm += g[i].d;
    }

    return 1.0 * sm / sum;
}

int main()
{
    while (scanf("%d%d",&n,&m) != EOF)
    {
        double minn = 10000000000;

        if (!m && !n) break;


        for (int i = 0;i < n;i++)
            scanf("%d",&b[i]);

        for (int i = 0;i < n;i++)
            for (int j = 0;j < n;j++)
            scanf("%d",&a[i][j]);

        int ba = (1 << n) - 1;

        for (int i = 3;i <= ba;i++)
        {
            int num = 0;

            for (int j = 0;j < n;j++)
            {
                if (i & (1 << j)) num++;
            }

            if (num == m)
            {
                //0printf("00
");
                double tmp = solve(i);

                if (tmp < minn)
                {
                    minn = tmp;
                }
            }
        }

        int mark;

        for (int i = 3;i <= ba;i++)
        {
            int num = 0;

            for (int j = 0;j < n;j++)
            {
                if (i & (1 << j)) num++;
            }

            if (num == m)
            {
                double tmp = solve(i);

                if (tmp == minn)
                {
                    mark = i;
                    break;
                }
            }
        }

        int p[20];

        int cnt = 0;

        for (int i = 0;i < n;i++)
        {
            if (mark & (1 << i)) p[cnt++] = i;
        }

        for (int i = 0;i < cnt;i++)
        {
            if (!i) printf("%d",p[i]+1);
            else printf(" %d",p[i]+1);
        }

        printf("
");
    }

    return 0;
}

再附上一份t了的代码：

#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <map>
#include <vector>
using namespace std;

int n,m;
int num = 0;
int par[20];
int a[20];
int b[20][20];
bool v[20];
int s[20];
vector<int> vv[10005];
map<double,int> mmp;

double minn;

struct node
{
    int x,y,d;
} g[500];

bool cmp(node aa,node bb)
{
    return aa.d < bb.d;
}

void init(int k)
{
    for (int i = 0;i <= k;i++)
        par[i] = i;
}

int fin(int x)
{
    if (x == par[x]) return x;
    else return par[x] = fin(par[x]);
}

void unit(int x,int y)
{
    x = fin(x);
    y = fin(y);

    if (x != y) par[x] = y;
}

void solve(void)
{
    int cnt = 0;

    bool f = 0;

    init(n);

    for (int i = 0;i < m;i++)
        for (int j = i + 1;j < m;j++)
    {
        int x = s[i],y = s[j];
        if (b[x][y])
        {
            g[cnt].x = x;
            g[cnt].y = y;
            g[cnt].d = b[x][y];
            cnt++;
        }
    }

    int sum = 0;

    for (int i = 0;i < m;i++)
    {
        int t = s[i];

        sum += a[t];
    }

    sort(g,g+cnt,cmp);

    int sm = 0;

    for (int i = 0;i < cnt;i++)
    {
        int x = g[i].x,y = g[i].y;

        if (fin(x) == fin(y)) continue;

        sm += g[i].d;

        unit(x,y);
    }

    int rt = fin(s[0]);

    for (int i = 0;i < m;i++)
    {
        if (fin(s[i]) != fin(rt)) f = 1;
    }

    double now = 1.0 * sm / sum;

    if (!f && now < minn)
    {
        if (!mmp[now])
        {
            mmp[now] = num;
            minn = now;

            for (int i = 0;i < m;i++)
                vv[num].push_back(s[i]);

            num++;
        }

    }
}

void dfs(int i,int p)
{
    if (m == n && p == m - 1)
    {
        s[p] = i;
        solve();
        return;
    }
    else if (m < n && p == m)
    {
        solve();
        return;
    }

    s[p] = i;

    for (int j = 0;j < n;j++)
    {
        if (!v[j])
        {
            v[j] = 1;
            dfs(j,p+1);
            v[j] = 0;
        }
    }
}


int main()
{
    while (scanf("%d%d",&n,&m) != EOF)
    {
        if (m == 0 && n == 0) break;

        num = 0;

        minn = 1000000000;

        memset(v,0,sizeof(v));
        memset(s,0,sizeof(s));
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
        memset(g,0,sizeof(g));
        memset(vv,0,sizeof(vv));

        for (int i = 0;i < n;i++)
            scanf("%d",&a[i]);


        for (int i = 0;i < n;i++)
            for (int j = 0;j < n;j++)
            scanf("%d",&b[i][j]);

        for (int i = 0;i < n;i++)
        {
            v[i] = 1;
            dfs(i,0);
            v[i] = 0;
        }

        int k = mmp[minn];

        sort(vv[k].begin(),vv[k].end());

        for (int i = 0;i < vv[k].size();i++)
        {
            if (!i) printf("%d",vv[k][i] + 1);
            else printf(" %d",vv[k][i] + 1);
        }

        printf("
");
    }

    return 0;
}