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            USACO chapter 2 section 2.4 Cow Tours

            USER: tian tianbing [tbbd4261]
            TASK: cowtour
            LANG: C++
            Compiling...
            Compile: OK
            Executing...
            Test 1: TEST OK [0.011 secs, 3212 KB]
            Test 2: TEST OK [0.000 secs, 3212 KB]
            Test 3: TEST OK [0.000 secs, 3212 KB]
            Test 4: TEST OK [0.000 secs, 3212 KB]
            Test 5: TEST OK [0.022 secs, 3212 KB]
            Test 6: TEST OK [0.022 secs, 3212 KB]
            Test 7: TEST OK [0.032 secs, 3212 KB]
            Test 8: TEST OK [0.032 secs, 3212 KB]
            Test 9: TEST OK [0.022 secs, 3212 KB]
            All tests OK.
            

            Your program ('cowtour') produced all correct answers! This is your submission #2 for this problem. Congratulations!

            /*
            ID:tbbd4261
            PROG:cowtour
            LANG:C++
            */

            #include
            <fstream>
            #include
            <iostream>
            #include
            <cmath>
            using namespace std;
            ifstream fin(
            "cowtour.in");
            ofstream fout(
            "cowtour.out");
            const int MAX=160;
            const double eps=1e-10, INT=1e30;

            double dist[MAX][MAX]={0};
            double dt[MAX]={0};
            int locate[MAX][2]={0};
            int n,i,j,k;
            char t;

            void Floyd()
            {
                 
            for(k=1; k<=n; k++)
                 
            for(i=1; i<=n; i++)
                 
            for(j=1; j<=n; j++)
                 {
                          
            if(dist[i][k]+dist[k][j]<dist[i][j])
                              dist[i][j]
            =dist[i][k]+dist[k][j];
                          
                 }
                 
            for(i=1; i<=n; i++)
                     dist[i][i]
            =INT;
            }

            int main()
            {
                fin
            >>n;
                
            for(i=1; i<=n; i++)
                         fin
            >>locate[i][0]>>locate[i][1];
                
            for(i=1; i<=n; i++)
                
            for(j=1; j<=n; j++)
                {
                         fin
            >>t;
                         t
            =t-'0';
                         
            if(t)dist[i][j]=sqrt( (locate[i][0]-locate[j][0])*(locate[i][0]-locate[j][0])+
                            ( locate[i][
            1]-locate[j][1])*(locate[i][1]-locate[j][1]) ) ;
                         
            else dist[i][j]=INT;
                }
                
                Floyd();
                
            double pmax=0,max=0,pmin=INT,tt;
                
            for(i=1; i<=n; i++)
                {
                     pmax
            =0;
                     
            for(j=1; j<=n; j++)
                          
            if(dist[i][j]>pmax&&dist[i][j]!=INT)pmax=dist[i][j];
                     dt[i]
            =pmax;
                     
            if(pmax>max)max=pmax;
                }
                
                
            for(i=1; i<=n-1; i++)
                
            for(j=i+1; j<=n; j++)
                {
                           
            if(dist[i][j]==INT&&i!=j)
                           {
                           tt
            =sqrt((locate[i][0]-locate[j][0])*(locate[i][0]-locate[j][0])+
                            ( locate[i][
            1]-locate[j][1])*(locate[i][1]-locate[j][1]) );
                           
            if(dt[i]+dt[j]+tt<pmin)pmin=dt[i]+dt[j]+tt;
                           }
                }
                fout.precision(
            6);
                fout
            <<fixed<<(pmin>max?pmin:max)<<endl;
                
            return 0;
            }

            posted on 2010-08-03 13:57 田兵 閱讀(197) 評論(0)  編輯 收藏 引用 所屬分類: USACO

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