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  • cogs 线型网络(状压dp)

    /*
    需要好大的空间..... 
    而且lowbit理解的不是很好
    先放到博客里 以后慢慢研究 
    */
    #include<iostream>
    #include<cstdio>
    #include<cstring>
    #include<cmath>
    #define inf 999999999;
    #define maxn 1050000
    using namespace std;
    int n,a[maxn];
    double x[30],y[30],g[30][30],f[maxn][21],ans=inf;
    double Dis(double x1,double y1,double x2,double y2)
    {
        return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
    }
    int l(int x)
    {
        return x&(-x);
    } 
    int main()
    {
        freopen("linec.in","r",stdin);
        freopen("linec.out","w",stdout);
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
          scanf("%lf%lf",&x[i],&y[i]);
        for(int i=1;i<=n;i++)
          for(int j=1;j<=n;j++)
            g[i][j]=Dis(x[i],y[i],x[j],y[j]);
        for(int i=0;i<=n;i++)
          a[1<<i]=i+1;
        for(int i=1;i<(1<<n);i++)if(!a[i])
          for(int j=i;j;j-=l(j))//枚举状态i的每一个1 
            {
              f[i][a[l(j)]]=inf;//更新此状态 
              for(int k=i-l(j);k;k-=l(k))
                f[i][a[l(j)]]=min(f[i][a[l(j)]],f[i-l(j)][a[l(k)]]+g[a[l(j)]][a[l(k)]]);
            }
        for(int i=1;i<=n;i++)
          ans=min(ans,f[(1<<n)-1][i]);
        printf("%.2f
    ",ans);
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/yanlifneg/p/5769562.html
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