两次DFS求树直径方法见 这里。
这里的直径是指最长链包含的节点个数，而上一题是指最长链的路径权值之和，注意区分。
K <= R: ans = K − 1;
K > R: ans = R − 1 + ( K − R ) ∗ 2;
1 #include <cstdio> 2 #include <cstring> 3 #include <cstdlib> 4 #include <algorithm> 5 6 using namespace std; 7 8 const int MAXN = 100010; 9 10 struct node 11 { 12 int v; 13 int next; 14 }; 15 16 int N, Q, EdgeN; 17 int head[MAXN]; 18 int best[MAXN]; 19 int dp[MAXN][3]; 20 node D[ MAXN << 1 ]; 21 bool vis[MAXN]; 22 23 void AddEdge( int u, int v ) 24 { 25 D[EdgeN].v = v; 26 D[EdgeN].next = head[u]; 27 head[u] = EdgeN++; 28 return; 29 } 30 31 void DFS1( int u ) 32 { 33 vis[u] = true; 34 for ( int i = head[u]; i != -1; i = D[i].next ) 35 { 36 int v = D[i].v; 37 int w = 1; 38 if ( !vis[v] ) 39 { 40 DFS1( v ); 41 if ( dp[v][0] + w > dp[u][0] ) 42 { 43 dp[u][1] = dp[u][0]; 44 dp[u][0] = dp[v][0] + w; 45 best[u] = v; 46 } 47 else if ( dp[v][0] + w > dp[u][1] ) 48 dp[u][1] = dp[v][0] + w; 49 } 50 } 51 return; 52 } 53 54 void DFS2( int u ) 55 { 56 vis[u] = true; 57 for ( int i = head[u]; i != -1; i = D[i].next ) 58 { 59 int fa = D[i].v; 60 int w = 1; 61 if ( !vis[fa] ) 62 { 63 dp[fa][2] = dp[u][2] + w; 64 if ( fa == best[u] ) 65 dp[fa][2] = max( dp[fa][2], dp[u][1] + w ); 66 else dp[fa][2] = max( dp[fa][2], dp[u][0] + w ); 67 DFS2( fa ); 68 } 69 } 70 return; 71 } 72 73 int main() 74 { 75 int T; 76 scanf( "%d", &T ); 77 while ( T-- ) 78 { 79 scanf( "%d%d", &N, &Q ); 80 EdgeN = 0; 81 memset( head, -1, sizeof(head) ); 82 for ( int i = 1; i < N; ++i ) 83 { 84 int u, v; 85 scanf( "%d%d", &u, &v ); 86 AddEdge( u, v ); 87 AddEdge( v, u ); 88 } 89 90 memset( dp, 0, sizeof(dp) ); 91 memset( vis, false, sizeof(bool) * (N + 1) ); 92 DFS1( 1 ); 93 memset( vis, false, sizeof(bool) * (N + 1) ); 94 DFS2( 1 ); 95 96 int maxx = 0; 97 for ( int i = 1; i <= N; ++i ) 98 maxx = max( maxx, max( dp[i][0], dp[i][2] ) ); 99 ++maxx; 100 101 while ( Q-- ) 102 { 103 int K; 104 scanf( "%d", &K ); 105 if ( maxx >= K ) 106 printf( "%d ", K - 1 ); 107 else printf( "%d ", maxx - 1 + ( K - maxx ) * 2 ); 108 } 109 } 110 return 0; 111 }