HDU-4607 Park Visit bfs | DP | dfs

　　题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4607

　　首先考虑找一条最长链长度k，如果m<=k+1，那么答案就是m。如果m>k+1，那么最长链上还有其他分支，来回走一遍，因此答案为2*m-k-1。。。求最长链可以DP，两次BFS或者DFS等。。

 //STATUS:C++_AC_453MS_3524KB
 #include <functional>
 #include <algorithm>
 #include <iostream>
 //#include <ext/rope>
 #include <fstream>
 #include <sstream>
 #include <iomanip>
 #include <numeric>
 #include <cstring>
 #include <cassert>
 #include <cstdio>
 #include <string>
 #include <vector>
 #include <bitset>
 #include <queue>
 #include <stack>
 #include <cmath>
 #include <ctime>
 #include <list>
 #include <set>
 #include <map>
 using namespace std;
 //using namespace __gnu_cxx;
 //define
 #define pii pair<int,int>
 #define mem(a,b) memset(a,b,sizeof(a))
 #define lson l,mid,rt<<1
 #define rson mid+1,r,rt<<1|1
 #define PI acos(-1.0)
 //typedef
 typedef __int64 LL;
 typedef unsigned __int64 ULL;
 //const
 const int N=;
 const LL INF=0x3f3f3f3f;
 const int MOD=,STA=;
 const LL LNF=1LL<<;
 const double EPS=1e-;
 const double OO=1e15;
 const int dx[]={-,,,};
 const int dy[]={,,,-};
 const int day[]={,,,,,,,,,,,,};
 //Daily Use ...
 inline int sign(double x){return (x>EPS)-(x<-EPS);}
 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
 template<class T> inline T lcm(T a,T b,T d){return a/d*b;}
 template<class T> inline T Min(T a,T b){return a<b?a:b;}
 template<class T> inline T Max(T a,T b){return a>b?a:b;}
 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}
 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}
 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}
 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}
 //End
 
 struct Edge{
     int u,v;
 }e[N*];
 int first[N],next[N*],d[N];
 int Ca,T,n,m,mt;
 
 void adde(int a,int b)
 {
     e[mt].u=a;e[mt].v=b;
     next[mt]=first[a];first[a]=mt++;
     e[mt].u=b;e[mt].v=a;
     next[mt]=first[b];first[b]=mt++;
 }
 
 int bfs(int s)
 {
     int u,i,hig;
     mem(d,);
     d[s]=;
     queue<int> q;
     q.push(s);
     hig=-;
     while(!q.empty())
     {
         u=q.front();q.pop();
         for(i=first[u];i!=-;i=next[i]){
             if(!d[e[i].v]){
                 d[e[i].v]=d[e[i].u]+;
                 if(d[e[i].v]>hig){
                     hig=d[e[i].v];
                     T=e[i].v;
                 }
                 q.push(e[i].v);
             }
         }
     }
     return hig;
 }
 
 int main()
 {
  //   freopen("in.txt","r",stdin);
     int i,j,a,b,hig;
     scanf("%d",&Ca);
     while(Ca--)
     {
         scanf("%d%d",&n,&m);
         mem(first,-);mt=;
         for(i=;i<n;i++){
             scanf("%d%d",&a,&b);
             adde(a,b);
         }
         bfs();
         hig=bfs(T);
 
         while(m--){
             scanf("%d",&a);
             if(a>hig){
                 printf("%d\n",*a-hig-);
             }
             else printf("%d\n",a-);
         }
     }
     return ;
 }