题意：给一个树编号0~n-1，一个数组a[i]为节点i在树上走的最大距离（不重复点），然后求最大的区间，使得区间最大差异小于某个值。dfs求出每个数组，同时区间化。枚举区间左边界，右边界同样递增，类似单调队列，区间最值用RMQ查询（常数小）。
1 #pragma comment(linker, "/STACK:10240000,10240000") 2 3 #include <iostream> 4 #include <cstdio> 5 #include <algorithm> 6 #include <cstdlib> 7 #include <cstring> 8 #include <map> 9 #include <queue> 10 #include <deque> 11 #include <cmath> 12 #include <vector> 13 #include <ctime> 14 #include <cctype> 15 #include <set> 16 17 using namespace std; 18 19 #define mem0(a) memset(a, 0, sizeof(a)) 20 #define lson l, m, rt << 1 21 #define rson m + 1, r, rt << 1 | 1 22 #define define_m int m = (l + r) >> 1 23 #define Rep(a, b) for(int a = 0; a < b; a++) 24 #define lowbit(x) ((x) & (-(x))) 25 #define constructInt4(name, a, b, c, d) name(int a = 0, int b = 0, int c = 0, int d = 0): a(a), b(b), c(c), d(d) {} 26 #define constructInt3(name, a, b, c) name(int a = 0, int b = 0, int c = 0): a(a), b(b), c(c) {} 27 #define constructInt2(name, a, b) name(int a = 0, int b = 0): a(a), b(b) {} 28 29 typedef double db; 30 typedef long long LL; 31 typedef pair<int, int> pii; 32 typedef multiset<int> msi; 33 typedef multiset<int>::iterator msii; 34 typedef set<int> si; 35 typedef set<int>::iterator sii; 36 typedef vector<int> vi; 37 38 const int dx[8] = {1, 0, -1, 0, 1, 1, -1, -1}; 39 const int dy[8] = {0, -1, 0, 1, -1, 1, 1, -1}; 40 const int maxn = 1e5 + 7; 41 const int maxm = 1e5 + 7; 42 const int maxv = 1e7 + 7; 43 const int MD = 1e9 +7; 44 const int INF = 1e9 + 7; 45 const double PI = acos(-1.0); 46 const double eps = 1e-10; 47 48 template<class T> struct MonotoneQueue{ 49 deque<T> Q; 50 MonotoneQueue<T>() { Q.clear(); } 51 void clear() { Q.clear(); } 52 bool empty() { return Q.empty(); } 53 void add_back(T x) { while (!Q.empty() && !(Q.back() < x)) Q.pop_back(); Q.push_back(x); } 54 void pop_front() { Q.pop_front(); } 55 T back2() { if(Q.size() < 2) return T(); return *(Q.end() - 2); } 56 T front() { return Q.front(); } 57 }; 58 59 template<class edge> struct Graph { 60 vector<vector<edge> > adj; 61 Graph(int n) { adj.clear(); adj.resize(n + 5); } 62 Graph() { adj.clear(); } 63 void resize(int n) { adj.resize(n + 5); } 64 void add(int s, edge e){ adj[s].push_back(e); } 65 void del(int s, edge e) { adj[s].erase(find(iter(adj[s]), e)); } 66 void clear() { adj.clear(); } 67 vector<edge>& operator [](int t) { return adj[t]; } 68 }; 69 70 Graph<int> G, W; 71 72 73 74 int maxd, id, d, n, m; 75 int dis[maxn], t[maxn], maxf[maxn][20], minf[maxn][20]; 76 bool vis[maxn]; 77 78 void DFS(int node) { 79 vis[node] = true; 80 dis[node] = max(dis[node], d); 81 if (d > maxd) { 82 maxd = d; 83 id = node; 84 } 85 for (int i = 0; i < G[node].size(); i++) { 86 int u = G[node][i]; 87 if (!vis[u]) { 88 d += W[node][i]; 89 DFS(u); 90 d -= W[node][i]; 91 } 92 } 93 } 94 95 void InitRMQ() { 96 for (int i = 1; i <= n; i++) maxf[i][0] = minf[i][0] = dis[i]; 97 for (int j = 1; (1 << j) <= n; j++) { 98 for (int i = 1; i + (1 << j) - 1 <= n; i++) { 99 maxf[i][j] = max(maxf[i][j - 1], maxf[i + (1 << (j - 1))][j - 1]); 100 minf[i][j] = min(minf[i][j - 1], minf[i + (1 << (j - 1))][j - 1]); 101 } 102 } 103 } 104 int RMQ_max(int L, int R) { 105 int x = t[R - L + 1]; 106 return max(maxf[L][x], maxf[R - (1 << x) + 1][x]); 107 } 108 int RMQ_min(int L, int R) { 109 int x = t[R - L + 1]; 110 return min(minf[L][x], minf[R - (1 << x) + 1][x]); 111 } 112 113 int solve(int q) { 114 int L = 1, ans = 0; 115 for (int R = 1; R <= n; R++) { 116 while (RMQ_max(L, R) - RMQ_min(L, R) > q) L++; 117 ans = max(ans, R - L + 1); 118 } 119 return ans; 120 } 121 122 int main() { 123 //freopen("in.txt", "r", stdin); 124 for (int i = 1; i <= 50000; i++) { 125 int j = 0; 126 while ((1 << (j + 1)) <= i) j++; 127 t[i] = j; 128 } 129 while (cin >> n >> m, n || m) { 130 G.clear(); G.resize(n); 131 W.clear(); W.resize(n); 132 for (int i = 1, u, v, w; i < n; i++) { 133 scanf("%d%d%d", &u, &v, &w); 134 G.add(u, v); 135 W.add(u, w); 136 G.add(v, u); 137 W.add(v, w); 138 } 139 mem0(vis); 140 maxd = -1; 141 DFS(1); 142 int node1 = id; 143 mem0(vis); 144 maxd = -1; 145 DFS(id); 146 int node2 = id; 147 mem0(dis); 148 mem0(vis); 149 DFS(node1); 150 mem0(vis); 151 DFS(node2); 152 InitRMQ(); 153 for (int i = 0, q; i < m; i++) { 154 scanf("%d", &q); 155 printf("%d ", solve(q)); 156 } 157 } 158 return 0; 159 }