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=100010;
const LL INF=0x3f3f3f3f;
const int MOD=1000000007,STA=8000010;
const LL LNF=1LL<<60;
const double EPS=1e-8;
const double OO=1e15;
const int dx[4]={-1,0,1,0};
const int dy[4]={0,1,0,-1};
const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
//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*2];
int first[N],next[N*2],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,0);
    d[s]=1;
    queue<int> q;
    q.push(s);
    hig=-1;
    while(!q.empty())
    {
        u=q.front();q.pop();
        for(i=first[u];i!=-1;i=next[i]){
            if(!d[e[i].v]){
                d[e[i].v]=d[e[i].u]+1;
                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,-1);mt=0;
        for(i=1;i<n;i++){
            scanf("%d%d",&a,&b);
            adde(a,b);
        }
        bfs(1);
        hig=bfs(T);

        while(m--){
            scanf("%d",&a);
            if(a>hig){
                printf("%d
",2*a-hig-1);
            }
            else printf("%d
",a-1);
        }
    }
    return 0;
}