hdu--4081--次小生成树<Kruskal--cool>

  1 #include <iostream>
  2 #include <cstring>
  3 #include <queue>
  4 #include <algorithm>
  5 #include <iomanip>
  6 #include <cmath>
  7 using namespace std;
  8 
  9 int cnt , num , n;
 10 const int size = 1010;
 11 
 12 int x[size];
 13 int y[size];
 14 int peo[size];//记录每个城市的人数
 15 int father[size];
 16 double maxDist[size][size];//记录从i->j这条路径上的边的最大值
 17 struct Tree//将所有2个城市之间的信息存入这里 st与end存入的是在0-n的当中编号 这边相当于将(x,y)这个坐标以一个点来表示
 18 {
 19     int st;
 20     int end;
 21     double dist;
 22     bool flag;
 23 }tree[size*size];
 24 struct graph //最小生成树的结点
 25 {
 26     int to;
 27     int next;
 28     double dist;
 29 }edge[size*2];
 30 int head[size];
 31 struct data//求次小生成树过程中需要的结点信息
 32 {
 33     int id;//当前结点
 34     double maxDist;//从遍历点i到当前结点的最大边的距离
 35     data( int u , double v):id(u),maxDist(v){};
 36     data(){};
 37 };
 38 bool cmp( const Tree p , const Tree q )
 39 {
 40     return p.dist < q.dist;//按边的权值从小到大进行排序--这边就是2点之间的距离大小
 41 }
 42 void init( )
 43 {
 44     memset( father , -1 , sizeof(father) );
 45     memset( head , -1 , sizeof(head) );
 46 }
 47 int find( int x )
 48 {
 49     return father[x] == -1 ? x : father[x] = find( father[x] );
 50 }
 51 void add1( int from , int to , double dist )
 52 {
 53     edge[cnt].to = to;
 54     edge[cnt].dist = dist;
 55     edge[cnt].next = head[from];
 56     head[from] = cnt++;
 57 }
 58 
 59 void add2( int st , int end , double dist )
 60 {
 61     tree[num].st = st;
 62     tree[num].end = end;
 63     tree[num].dist = dist;
 64     tree[num++].flag = false;
 65 }
 66 
 67 double getDist( int p , int q )
 68 {
 69     return sqrt( ( (x[p]-x[q])*(x[p]-x[q]) + (y[p]-y[q])*(y[p]-y[q])*1.0 ) );
 70 }
 71 
 72 double kruskal( )
 73 {
 74     init();
 75     int x , y;
 76     int edgeNum = 0;
 77     double ans = 0;
 78     sort( tree , tree+num , cmp );
 79     for( int i = 0 ; i<num ; i++ )//这里的i都是指第 I 条边 
 80     {
 81         x = find( tree[i].st );
 82         y = find( tree[i].end );
 83         if(x!=y)//如果这条边的两个端点不属于一个连通块 就可以进行合并
 84         {
 85             edgeNum ++;
 86             add1( tree[i].st,tree[i].end,tree[i].dist );
 87             add1( tree[i].end,tree[i].st,tree[i].dist );
 88             tree[i].flag = true;//加入mst之中
 89             father[x] = y;
 90             ans += tree[i].dist;
 91         }
 92         if( edgeNum == n-1 )
 93             break;
 94     }
 95     return edgeNum == n-1 ? ans : -1;
 96 }
 97 
 98 void bfs( int p )
 99 {
100     data now;
101     bool vis[size];
102     memset( vis , false , sizeof(vis) );
103     queue<data>q;
104     q.push( data(p,0) );
105     vis[p] = true;
106     while( !q.empty() )
107     {
108         now = q.front();
109         q.pop();
110         for( int i = head[now.id] ; i!=-1 ; i = edge[i].next )
111         {
112             int end = edge[i].to;
113             double dist = edge[i].dist;
114             if( !vis[end] )
115             {
116                 if( now.maxDist>dist )
117                     dist = now.maxDist;
118                 maxDist[p][end] = dist;
119                 vis[end] = true;
120                 q.push( data(end,dist) );
121             }
122         }
123     }
124 }
125 
126 void secondMst( double mst )
127 {
128     double ans = -1;
129     for( int i = 0 ; i<n ; i++ )
130     {
131         bfs( i );
132     }    
133     for( int i = 0 ; i<num ; i++ )//num遍历所有的边-尝试将该边设置为magic road--根据该边是否属于mst集合 进行不同操作
134     {
135         double temp;
136         if( !tree[i].flag )//该边不属于mst集合之中
137         {
138             temp = ( peo[ tree[i].st ] + peo[ tree[i].end ] )*1.0 / ( mst - maxDist[ tree[i].st ][ tree[i].end ] );  
139         }
140         else
141         {
142             temp = ( peo[ tree[i].st ] + peo[ tree[i].end ] )*1.0 / ( mst - tree[i].dist );
143         }
144         ans = max( ans , temp );
145     }
146     cout<<setiosflags(ios::fixed);
147     cout<<setprecision(2)<<ans<<endl;
148 }
149 
150 int main()
151 {
152     cin.sync_with_stdio(false);
153     int t;
154     cin >> t;
155     while( t-- )
156     {
157         num = cnt = 0;
158         cin >> n;
159         for( int i = 0 ; i<n ; i++ )
160         {
161             cin >> x[i] >> y[i] >> peo[i];
162         }
163         for( int i = 0 ; i<n ; i++ )
164         {
165             for( int j = 0 ; j<i ; j++ )
166             {
167                 double val = getDist(i,j);
168                 add2(i,j,val);
169             }
170         }
171         double mst = kruskal();
172         secondMst( mst );
173     }
174     return 0;
175 }