loj 1412(树上最长直径的应用)

题目链接：http://lightoj.com/volume_showproblem.php?problem=1412

思路：好久没写题解了，有点手生，这题从昨天晚上wa到现在终于是过了。。。思想其实很简单，就是预处理出每一块的最长直径，然后每次询问的时候直接查询就可以了。

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<vector>
 using namespace std;

 const int MAXN = ( + );
 typedef pair<int,int>Pair;

 vector<int >g[MAXN];
 vector<Pair >blocks;
 int n,m,q,len,_count,max_len;
 bool mark[MAXN];

 int dfs(int u,int father)
 {
     if(!mark[u])mark[u] = true, _count ++;
     int first = , second = ;
     for(int i = ; i < (int)g[u].size(); i ++) {
         int v = g[u][i];
         if(v == father)continue;
         int tmp = dfs(g[u][i],u) + ;
         if(tmp > first)second = first, first = tmp;
         else if(tmp > second)second = tmp;
     }
     if(first + second > len)len = first + second;
     return first;
 }

 int cmp(Pair p, Pair q)
 {
     if(p.second != q.second){
         return p.second > q.second;
     }else
         return p.first > q.first;
 }

 int main()
 {
     int u,v,_case,t=;
     scanf("%d",&_case);
     while(_case--){
         scanf("%d%d",&n,&m);
         for(int i=;i<=n;i++)g[i].clear();
         while(m--){
             scanf("%d%d",&u,&v);
             g[u].push_back(v);
             g[v].push_back(u);
         }
         len=;
         max_len = ;
         memset(mark,false,sizeof(mark));
         blocks.clear();
         for(int i=;i<=n;i++){
             if(!mark[i]){
                 len=;
                 _count = ;
                 dfs(i,-);
                 blocks.push_back(make_pair(len,_count));
                 max_len = max(max_len,len);
             }
         }
         sort(blocks.begin(),blocks.end(),cmp);
         scanf("%d",&q);
         printf("Case %d:\n",t++);
         while(q--){
             int k;
             scanf("%d",&k);
             if(k > blocks[].second){
                 puts("impossible");
             }else if(k <= max_len + ){
                 printf("%d\n",k - );
             }else {
                 int ans = ( << );
                 for(int i=; i<(int)blocks.size(); i++){
                     if(k > blocks[i].second)break;
                     ans = min (ans, (blocks[i].first)+*(k-blocks[i].first-));
                 }
                 printf("%d\n",ans);
             }
         }

     }
     return ;
 }