USACO Section 2.4 Cow Tours - 考虑要全面阿...

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这道题最值得注意的是牧场数>=2...也就是说把两个联通图相连后得到的新牧场大小并不一定是最大的..可能还有另外一个没连的直径更长...

其他的..用并查集来判断是否在一个联通图里..用Floyd求两点的最短路径~~

进入第三章了~~

Program:

/*  ID: zzyzzy12  LANG: C++  TASK: cowtour*/    #include<iostream>    #include<istream>#include<stdio.h>    #include<string.h>    #include<math.h>    #include<stack>#include<algorithm>    #include<queue> #define oo 2000000000using namespace std;struct node{     int y,x;       }point[155];double p[155][155],ans,ans2,sum[155]; int n,father[155];bool getdata(){     char c=' ';      while (c!='1' && c!='0') c=getchar();     if (c=='1') return true;     return false;}double dis(node a,node b){     return  sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));     }int getfather(int i){     if (father[i]==i) return i;     return father[i]=getfather(father[i]);   }void Floyd(){     int k,i,j;     for (k=1;k<=n;k++)        for (i=1;i<=n;i++)           for (j=1;j<=n;j++)                if (p[i][j]>p[i][k]+p[k][j])                 p[i][j]=p[i][k]+p[k][j];     return;    } int main(){     freopen("cowtour.in","r",stdin);       freopen("cowtour.out","w",stdout);      scanf("%d",&n);     int i,j,k;      for (i=1;i<=n;i++)      {           scanf("%d%d",&point[i].x,&point[i].y);           p[i][i]=0;           father[i]=i;     }     for (i=1;i<=n;i++)        for (j=1;j<=n;j++)          if (getdata())           {                p[i][j]=dis(point[i],point[j]);                father[getfather(i)]=getfather(j);          }          else if (i!=j) p[i][j]=oo;     Floyd();     memset(sum,0,sizeof(sum));     for (i=1;i<=n;i++)        for (j=1;j<=n;j++)          if (getfather(i)==getfather(j) && p[i][j]>sum[i])             sum[i]=p[i][j];     ans2=0;     for (i=1;i<=n;i++)        if (ans2<sum[i]) ans2=sum[i];     ans=oo;     for (i=1;i<=n;i++)        for (j=1;j<=n;j++)          if (father[i]!=father[j] && ans>sum[i]+sum[j]+dis(point[i],point[j]))             ans=sum[i]+sum[j]+dis(point[i],point[j]);     if (ans2>ans) ans=ans2;     printf("%.6lf\n",ans);     return 0;   }