hdu 4607 Park Visit (dfs)
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4607
首先如果k小于等于直径长度,那么答案为k−1。如果k大于直径长度,设直径长度为r,那么答案为r−1+(k−r)∗2。
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1 #include <cstdio> 2 #include <cmath> 3 #include <algorithm> 4 #include <iostream> 5 #include <cstring> 6 #include <queue> 7 #include <vector> 8 #define maxn 105000 9 using namespace std; 10 11 const int INF = 0x3f3f3f; 12 13 vector<int> G[maxn]; 14 int N,M; 15 int maxdeep,point; 16 bool vis[maxn]; 17 void dfs(int u,int deep){ 18 if(maxdeep < deep){ 19 maxdeep = deep; 20 point = u; 21 } 22 for(int i=0;i<G[u].size();i++){ 23 int v = G[u][i]; 24 if(!vis[v]){ 25 vis[v] = true; //printf("u,v %d %d\n",u,v);jf 26 dfs(v,deep+1); 27 } 28 } 29 } 30 31 int main() 32 { 33 //if(freopen("input.txt","r",stdin)== NULL) {printf("Error\n"); exit(0);} 34 35 int T; 36 cin>>T; 37 while(T--){ 38 cin>>N>>M; 39 if(N == 1){ 40 for(int i=1;i<=M;i++){ 41 int K; 42 scanf("%d",&K); 43 printf("0\n"); 44 } 45 continue; 46 } 47 48 for(int i=1;i<=N;i++) G[i].clear(); 49 for(int i=1;i<=N-1;i++){ 50 int a,b; 51 scanf("%d %d",&a,&b); 52 G[a].push_back(b); 53 G[b].push_back(a); 54 } 55 memset(vis,0,sizeof(vis)); 56 maxdeep = 0; point = 0; 57 vis[1] = true; 58 dfs(1,0); 59 memset(vis,0,sizeof(vis)); 60 dfs(point,0); 61 for(int i=1;i<=M;i++){ 62 int K; 63 scanf("%d",&K); 64 if(maxdeep >= K-1) printf("%d\n",K-1); 65 else printf("%d\n",maxdeep + 2 * (K - 1 - maxdeep)); 66 } 67 } 68 }