TOJ--2119--最小生成树和map

这题唯一的价值应该就是稍微用了下map 同时也算自己对于prim算法的再次练手吧.....

其余的没什么好讲的就是保留1位小数这边的数据范围题目没有给出我也一直不知道......明天考6J了.....

说些什么上帝才能听到我的祈求呢~

  1 // TOJ 2119 最小生成树
  2 
  3 #include <iostream>
  4 #include <cstdio>
  5 #include <cstring>
  6 #include <map>
  7 using namespace std;
  8 
  9 const int size = 520;
 10 const double inf = 888888888.0;
 11 double length;
 12 int n;
 13 string name;
 14 struct data
 15 {
 16     int num;
 17     int next[50];
 18     double dist[50];
 19 }edge[size];
 20 map<string,int>mp;
 21 bool vis[size];
 22 double dist[size];
 23 
 24 void prim()
 25 {
 26     int i , j , k;
 27     memset( vis , false , sizeof(false) );
 28     for( i = 0 ; i<size ; i++ )
 29     {
 30         dist[i] = (i==0)?0:inf;
 31     }
 32     for( i = 0 ; i<edge[0].num ; i++ )
 33     {
 34         dist[ edge[0].next[i] ] = edge[0].dist[i];
 35     }
 36     vis[0] = true;
 37     double ans = 0;
 38     double mmin;
 39     for( i = 1 ; i<n ; i++ )
 40     {
 41         mmin = inf;
 42         for( j = 1 ; j<n ; j++ )
 43         {
 44             if( !vis[j] && dist[j]<mmin )
 45             {
 46                 mmin = dist[j];
 47                 k = j;
 48             }
 49         }
 50         if( mmin == inf )
 51         {
 52             break;
 53         }
 54         vis[k] = true;
 55         ans+=mmin;
 56         for( j = 0 ; j<edge[k].num ; j++ )
 57         {
 58             if( dist[ edge[k].next[j] ] > edge[k].dist[j]  && !vis[ edge[k].next[j] ]  )
 59             {
 60                 dist[ edge[k].next[j] ] = edge[k].dist[j];
 61             }
 62         }
 63     }
 64     if( ans<=length )
 65         printf( "Need %.1lf miles of cable
",ans );
 66     else
 67         printf( "Not enough cable
" );
 68 }
 69 
 70 int main()
 71 {
 72     char st[30] , end[30];
 73     double len;        
 74     int m;
 75     int cnt;
 76     while( ~scanf("%lf",&length) )
 77     {
 78         mp.clear();
 79         for( int i = 0 ; i<size ; i++ )
 80             edge[i].num = 0;
 81         cnt = 0;
 82         scanf( "%d",&n );
 83         for ( int i = 0 ; i<n ; i++ )
 84         {
 85             cin>>name;
 86             mp[name] = cnt++;  // 将人名映射为数字
 87         }
 88         scanf( "%d",&m );
 89         while( m-- )
 90         {
 91             scanf( "%s %s %lf",st,end,&len );
 92             int x = mp[st];
 93             int y = mp[end];
 94             edge[x].next[ edge[x].num ] = y;
 95             edge[x].dist[ edge[x].num++ ] = len;
 96             edge[y].next[ edge[y].num ] = x;
 97             edge[y].dist[ edge[y].num++ ] = len;
 98         }
 99         prim();
100     }
101     return 0;
102 }

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