HDU-4607 Park Visit bfs | DP | dfs - Linux

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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 ;

 }