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。。
1 //STATUS:C++_AC_453MS_3524KB 2 #include <functional> 3 #include <algorithm> 4 #include <iostream> 5 //#include <ext/rope> 6 #include <fstream> 7 #include <sstream> 8 #include <iomanip> 9 #include <numeric> 10 #include <cstring> 11 #include <cassert> 12 #include <cstdio> 13 #include <string> 14 #include <vector> 15 #include <bitset> 16 #include <queue> 17 #include <stack> 18 #include <cmath> 19 #include <ctime> 20 #include <list> 21 #include <set> 22 #include <map> 23 using namespace std; 24 //using namespace __gnu_cxx; 25 //define 26 #define pii pair<int,int> 27 #define mem(a,b) memset(a,b,sizeof(a)) 28 #define lson l,mid,rt<<1 29 #define rson mid+1,r,rt<<1|1 30 #define PI acos(-1.0) 31 //typedef 32 typedef __int64 LL; 33 typedef unsigned __int64 ULL; 34 //const 35 const int N=100010; 36 const LL INF=0x3f3f3f3f; 37 const int MOD=1000000007,STA=8000010; 38 const LL LNF=1LL<<60; 39 const double EPS=1e-8; 40 const double OO=1e15; 41 const int dx[4]={-1,0,1,0}; 42 const int dy[4]={0,1,0,-1}; 43 const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31}; 44 //Daily Use ... 45 inline int sign(double x){return (x>EPS)-(x<-EPS);} 46 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;} 47 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;} 48 template<class T> inline T lcm(T a,T b,T d){return a/d*b;} 49 template<class T> inline T Min(T a,T b){return a<b?a:b;} 50 template<class T> inline T Max(T a,T b){return a>b?a:b;} 51 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);} 52 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);} 53 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));} 54 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));} 55 //End 56 57 struct Edge{ 58 int u,v; 59 }e[N*2]; 60 int first[N],next[N*2],d[N]; 61 int Ca,T,n,m,mt; 62 63 void adde(int a,int b) 64 { 65 e[mt].u=a;e[mt].v=b; 66 next[mt]=first[a];first[a]=mt++; 67 e[mt].u=b;e[mt].v=a; 68 next[mt]=first[b];first[b]=mt++; 69 } 70 71 int bfs(int s) 72 { 73 int u,i,hig; 74 mem(d,0); 75 d[s]=1; 76 queue<int> q; 77 q.push(s); 78 hig=-1; 79 while(!q.empty()) 80 { 81 u=q.front();q.pop(); 82 for(i=first[u];i!=-1;i=next[i]){ 83 if(!d[e[i].v]){ 84 d[e[i].v]=d[e[i].u]+1; 85 if(d[e[i].v]>hig){ 86 hig=d[e[i].v]; 87 T=e[i].v; 88 } 89 q.push(e[i].v); 90 } 91 } 92 } 93 return hig; 94 } 95 96 int main() 97 { 98 // freopen("in.txt","r",stdin); 99 int i,j,a,b,hig; 100 scanf("%d",&Ca); 101 while(Ca--) 102 { 103 scanf("%d%d",&n,&m); 104 mem(first,-1);mt=0; 105 for(i=1;i<n;i++){ 106 scanf("%d%d",&a,&b); 107 adde(a,b); 108 } 109 bfs(1); 110 hig=bfs(T); 111 112 while(m--){ 113 scanf("%d",&a); 114 if(a>hig){ 115 printf("%d\n",2*a-hig-1); 116 } 117 else printf("%d\n",a-1); 118 } 119 } 120 return 0; 121 }
