USACO 2.4 Cow Tours

TASK: cowtour
LANG: C++

Compiling...
Compile: OK

Executing...
   Test 1: TEST OK [0.000 secs, 3256 KB]
   Test 2: TEST OK [0.000 secs, 3256 KB]
   Test 3: TEST OK [0.000 secs, 3256 KB]
   Test 4: TEST OK [0.000 secs, 3256 KB]
   Test 5: TEST OK [0.000 secs, 3256 KB]
   Test 6: TEST OK [0.000 secs, 3256 KB]
   Test 7: TEST OK [0.000 secs, 3256 KB]
   Test 8: TEST OK [0.000 secs, 3256 KB]
   Test 9: TEST OK [0.000 secs, 3256 KB]

All tests OK.
  1 /*
  2 PROG: cowtour
  3 ID: jiafeim1
  4 LANG: C++
  5 */
  6 
  7 
  8 #include <iostream>
  9 #include <fstream>
 10 #include <string>
 11 #include <iomanip>
 12 using namespace std;
 13 
 14 #define sq(x) (((double)x)*(x))
 15 #define maxD(x,y) ((x)>(y)?(x):(y))
 16 int n;
 17 #include <cmath>
 18 
 19 long point[160][3];
 20 bool linjie[160][160] = {false};
 21 double len[160][160];
 22 double longest[160]={0.0};
 23 double filed_longest[160]={0.0};
 24 
 25 int main()
 26 {
 27     ifstream fin("cowtour.in");
 28     ofstream fout("cowtour.out");
 29 
 30     fin>>n;
 31 
 32     for(int i = 0 ;i!=n;++i)
 33     {
 34         fin>>point[i][0]>>point[i][1];
 35     }
 36 
 37     for(int i = 0;i!=n;++i)
 38     {
 39         for(int j = 0 ;j!=n;++j)
 40         {
 41             if(i!=j)
 42                 len[i][j] = 999999999999.0;
 43             else
 44                 len[i][j] = 0.0;
 45         }
 46     }
 47     char temp;
 48     for(int i = 0;i!=n;++i)
 49     {
 50         for(int j = 0 ;j!=n;++j)
 51         {
 52             fin>>temp;
 53             if(temp == '1')
 54             {
 55                 linjie[i][j] = true;
 56                 if(!linjie[j][i])
 57                     len[j][i] = len[i][j] = sqrt(sq(point[i][0]-point[j][0]) + sq(point[i][1]-point[j][1]));
 58             }
 59         }
 60     }
 61     for(int w = 0;w!=n;++w)
 62         for(int i = 0;i!=n;++i)
 63             for(int j = 0 ;j!=n;++j)          
 64             {
 65                 if(len[i][j]>len[i][w]+len[w][j])
 66                     len[i][j] = len[i][w]+len[w][j];
 67             }
 68     for(int i = 0;i!=n;++i)
 69         for(int j = 0 ;j!=n;++j)
 70         {
 71             if(longest[i]<len[i][j] && len[i][j]<99999999999.0)
 72                 longest[i]=len[i][j];
 73         }
 74     
 75     for(int i = 0;i!=n;++i)
 76         for(int j = 0 ;j!=n;++j)
 77         {
 78             if(len[i][j]<99999999999.0 && filed_longest[i]<longest[j])
 79                 filed_longest[i] = longest[j];
 80         }
 81 
 82 
 83     double minD = 999999999999.0;
 84     for(int i = 0;i!=n;++i)
 85         for(int j = 0 ;j!=n;++j)
 86         {
 87             if(len[i][j]>99999999999.0)
 88             {
 89                 double l = longest[i]+longest[j]+sqrt(sq(point[i][0]-point[j][0]) + sq(point[i][1]-point[j][1]));
 90                 double res = maxD(maxD(l,filed_longest[i]),filed_longest[j]);
 91                 if(res<minD) minD = res;
 92             }
 93         }
 94     
 95     
 96     fout << fixed << setprecision(6) << minD << endl;
 97     
 98     fin.close();
 99     fout.close();
100     
101     return 0;
102 }
posted @ 2011-05-08 10:58  幻魇  阅读(335)  评论(0)  编辑  收藏  举报