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代码：
1 #include <bits/stdc++.h> 2 3 using namespace std; 4 5 #define re register 6 #define rep(i, a, b) for (re int i = a; i <= b; ++i) 7 #define repd(i, a, b) for (re int i = a; i >= b; --i) 8 #define For(i, a, b, s) for (re int i = a; i <= b; s) 9 #define maxx(a, b) a = max(a, b) 10 #define minn(a, b) a = min(a, b) 11 #define LL long long 12 #define INF (1 << 30) 13 14 inline int read() { 15 int w = 0, f = 1; char c = getchar(); 16 while (!isdigit(c)) f = c == '-' ? -1 : f, c = getchar(); 17 while (isdigit(c)) w = (w << 3) + (w << 1) + (c ^ '0'), c = getchar(); 18 return w * f; 19 } 20 21 const int maxn = 1e4 + 5, maxm = 100 + 5; 22 23 struct Edge { 24 int u, v, w, pre; 25 } e[maxn << 1]; 26 int ec, G[maxn]; 27 void init() { memset(G, -1, sizeof(G)); ec = 0; } 28 void add(int u, int v, int w) { e[ec++] = (Edge){u, v, w, G[u]}; G[u] = ec-1; } 29 #define iter(i, u) for (register int i = G[u]; i != -1; i = e[i].pre) 30 31 struct Pair { 32 int d, id; 33 bool operator < (const Pair &r) const { return d < r.d; } 34 } p[maxn]; 35 int N, M; 36 int ask[maxm], ans[maxm]; 37 int son[maxn], rt, maxp[maxn], vis[maxn], cnt; 38 39 void get_dist(int u, int fa, int rt, int dist) { 40 son[u] = 1; p[cnt++] = (Pair){dist, rt}; 41 iter(i, u) 42 if (e[i].v != fa && !vis[e[i].v]) { 43 get_dist(e[i].v, u, rt, dist + e[i].w); 44 son[u] += son[e[i].v]; 45 } 46 } 47 48 int query(int v) { 49 int l = 0, r = cnt-1, res = cnt; 50 while (l <= r) { 51 int mid = (l + r) >> 1; 52 if (v >= p[mid].d) l = mid+1, res = mid; 53 else r = mid-1; 54 } 55 return res; 56 } 57 58 void solve(int u) { 59 cnt = 1; 60 iter(i, u) { 61 if (vis[e[i].v]) continue; 62 get_dist(e[i].v, u, e[i].v, e[i].w); 63 } 64 sort(p, p+cnt); 65 rep(i, 1, M) if (!ans[i]) 66 rep(s, 0, cnt) { 67 int t = query(ask[i]-p[s].d); 68 while (p[s].id == p[t].id && p[t].d == p[t+1].d) t++; 69 if (s == t || p[s].id == p[t].id) continue; 70 if (p[s].d + p[t].d == ask[i]) { 71 ans[i] = 1; break; 72 } 73 } 74 } 75 76 int get_rt(int u, int fa, int sum) { 77 maxp[u] = 0; int son = 1; 78 iter(i, u) 79 if (e[i].v != fa && !vis[e[i].v]) { 80 int child = get_rt(e[i].v, u, sum); 81 maxx(maxp[u], child); 82 son += child; 83 } 84 maxx(maxp[u], sum - son); 85 if (maxp[u] < maxp[rt]) rt = u; 86 return son; 87 } 88 89 void divid(int u) { 90 solve(u); vis[u] = 1; 91 iter(i, u) 92 if (!vis[e[i].v]) { 93 rt = 0; 94 get_rt(e[i].v, e[i].v, son[e[i].v]); 95 divid(rt); 96 } 97 } 98 99 int main() { 100 init(); maxp[rt = 0] = INF; 101 N = read(), M = read(); 102 rep(i, 1, N-1) { 103 int u = read(), v = read(), w = read(); 104 add(u, v, w); add(v, u, w); 105 } 106 rep(i, 1, M) ask[i] = read(); 107 rt = 0; get_rt(1, 1, N); divid(rt); 108 rep(i, 1, M) printf("%s\n", ans[i] ? "AYE" : "NAY"); 109 return 0; 110 }
