hdu 3721 树的最小直径

题意：
      给你一棵树，让你改变一条边，改变之后依然是一棵树，然后问你怎样改变才能让树的直径最短。这里的改变一条边指的是指把一条边长度不变，连在别的两个点上。


思路：

      首先求出树的直径，把直径上的边记录下来，然后在枚举这些边（枚举别的边没意义）每次枚举我的做法是后建造两棵树，我们只要在这两棵树之间连接一条边就行了，但是怎么连接呢？ 我是先没别求两棵树的直径，然后在找到直径上中间点，然后连接这两棵树的中间点，只有这样才能保证最短，每次连接后的直径就是 两棵树的直径，和当前枚举的边长度+两个树被中间点分开的较长的那一个值的和，他们三个中较长的哪一个，就这样在所有中找到一个最小的就是答案。

#include<stdio.h>
#include<string.h>
#include<queue>
#include<map>

#define N_node 2500 + 5
#define N_edge 5000 + 5
#define INF 1000000000

using namespace std;

typedef struct
{
   int to ,next ,cost;
}STAR;

typedef struct
{
   int a ,b ,c;
}EDGE;

STAR E[N_edge];
EDGE edge[N_node];
int list[N_node] ,tot;
int s_x[N_node] ,mer[N_node];
map<int ,map<int ,int> >hash;

void add(int a ,int b ,int c)
{
   E[++tot].to = b;
   E[tot].cost = c;
   E[tot].next = list[a];
   list[a] = tot;
   
   E[++tot].to = a;
   E[tot].cost = c;
   E[tot].next = list[b];
   list[b] = tot;
}

void Spfa(int s ,int n)
{
   for(int i = 0 ;i <= n ;i ++)
   s_x[i] = INF ,mer[i] = i;
   int mark[N_node] = {0};
   s_x[s] = 0 ,mark[s] = 1;
   queue<int>q;
   q.push(s);   
   while(!q.empty())
   {
      int xin ,tou;
      tou = q.front();
      q.pop();
      mark[tou] = 0;
      for(int k = list[tou] ;k ;k = E[k].next)
      {
         xin = E[k].to;
         if(s_x[xin] > s_x[tou] + E[k].cost)
         {
            s_x[xin] = s_x[tou] + E[k].cost;
            mer[xin] = tou;
            if(!mark[xin])
            {
               mark[xin] = 1;
               q.push(xin);
            }
         }
      }
   }
   return ;
}

int abss(int x)
{
   return x > 0 ? x : -x;
}

int maxx(int x ,int y)
{
   return x > y ? x : y;
}

int main ()
{
   int t ,n ,a ,b ,c ,i ,j;
   int cas  = 1;
   scanf("%d" ,&t);
   while(t--)
   {
      scanf("%d" ,&n);
      memset(list ,0 ,sizeof(list));
      tot = 1;
      for(i = 1 ;i < n ;i ++)
      {
         scanf("%d %d %d" ,&edge[i].a ,&edge[i].b ,&edge[i].c);
         edge[i].a ++ ,edge[i].b ++;
         add(edge[i].a ,edge[i].b ,edge[i].c);
      }
      
      int p01 ,p02;
      Spfa(1 ,n);   
      int maxxx = -1;
      for(j = 1 ;j <= n ;j ++)
      if(maxxx < s_x[j] && s_x[j] != INF)
      {
         maxxx = s_x[j];
         p01 = j;
      }
      Spfa(p01 ,n);  
      maxxx = -1;
      for(j = 1 ;j <= n ;j ++)
      if(maxxx < s_x[j] && s_x[j] != INF)
      {
         maxxx = s_x[j];
         p02 = j;
      }
      int x = p02;
      hash.clear();
      while(x != mer[x])
      {
         hash[x][mer[x]] = hash[mer[x]][x] = 1;    
         x = mer[x];
      }
      
      
        
      int ans = INF;          
      for(i = 1 ;i < n ;i ++)
      {
         if(!hash[edge[i].a][edge[i].b]) continue;
         memset(list ,0 ,sizeof(list)) ,tot = 1;
         for(j = 1 ;j < n ;j ++)
         if(j == i) continue;
         else add(edge[j].a ,edge[j].b ,edge[j].c);
         
         int p11 ,p12 ,mid1 ,l1 ,mid1l;
         Spfa(edge[i].a ,n);   
         int max = -1;
         for(j = 1 ;j <= n ;j ++)
         if(max < s_x[j] && s_x[j] != INF)
         {
            max = s_x[j];
            p11 = j;
         }
         Spfa(p11 ,n);  
         max = -1;
         for(j = 1 ;j <= n ;j ++)
         if(max < s_x[j] && s_x[j] != INF)
         {
            max = s_x[j];
            p12 = j;
         }
         l1 = max;
         int x = p12;
         int min = INF;
         while(x != mer[x])
         {
            if(min > abss(s_x[x] - (l1 - s_x[x])))
            {
               min = abss(s_x[x] - (l1 - s_x[x]));
               mid1 = x;
               mid1l = maxx(s_x[x] ,l1 - s_x[x]);
            }
            x = mer[x];
         }  
         if(min > abss(s_x[x] - (l1 - s_x[x])))
         {
            min = abss(s_x[x] - (l1 - s_x[x]));
            mid1 = x;
            mid1l = maxx(s_x[x] ,l1 - s_x[x]);
         }
         
         
         int p21 ,p22 ,mid2 ,l2 ,mid2l;
         Spfa(edge[i].b ,n); 
         max = -1;
         for(j = 1 ;j <= n ;j ++)
         if(max < s_x[j] && s_x[j] != INF)
         {
            max = s_x[j];
            p21 = j;
         }
         Spfa(p21 ,n);
         max = -1;
         for(j = 1 ;j <= n ;j ++)
         if(max < s_x[j] && s_x[j] != INF)
         {
            max = s_x[j];
            p22 = j;
         }
         l2 = max;
         x = p22;
         min = INF;   
         while(x != mer[x])
         {
            if(min > abss(s_x[x] - (l2 - s_x[x])))
            {
               min = abss(s_x[x] - (l2 - s_x[x]));
               mid2 = x;
               mid2l = maxx(s_x[x] ,l2 - s_x[x]);
            }
            x = mer[x];
         }  
         if(min > abss(s_x[x] - (l2 - s_x[x])))
         {
            min = abss(s_x[x] - (l2 - s_x[x]));
            mid2 = x;
            mid2l = maxx(s_x[x] ,l2 - s_x[x]);
         }
         
         int now = maxx(maxx(l1 ,l2) ,edge[i].c + mid1l + mid2l);
         if(ans > now) ans = now;
      }
      printf("Case %d: %d
" ,cas ++ ,ans);
   }
   return 0;
}