题意：
给定一颗树，求一个字典序最小的路径，路径的权值是点权的累成mod(1e6+3)，要求点权的乘积为k。
题解：
树上路径，用点分治可以做。按照点分治的套路来，x * y = k 那么枚举x = k * inv[y] % mod，为刚开始用来set来求最小，被t飞了， 看了题解发现可以用树组优化掉一个log。
1 #include <bits/stdc++.h> 2 using namespace std; 3 #define N 100010 4 #define M 1000010 5 #define mod 1000003 6 typedef long long ll; 7 typedef pair<int, int> PII; 8 inline int read() 9 { 10 char ch = getchar();int x = 0, f = 1; 11 while(ch < '0' || ch > '9') {if(ch == '-') f = -1; ch = getchar();} 12 while(ch >= '0' && ch <= '9') {x = (x << 1) + (x << 3) - '0' + ch; ch = getchar();} 13 return x * f; 14 } 15 16 int n, k, size[N], rt, tot, st[N], mx[N],a[N]; 17 18 int ans1, ans2; 19 int inv[M]; 20 int h[N], e[N * 2], ne[N * 2], idx; 21 ll tmp[N], top, id[N]; 22 ll mp[M]; 23 24 void add(int a, int b) 25 { 26 e[idx] = b; 27 ne[idx] = h[a]; 28 h[a] = idx ++; 29 } 30 31 void getrt(int u, int fa) 32 { 33 mx[u] = 0; size[u] = 1; 34 for (int i = h[u]; ~i; i = ne[i]) 35 { 36 int j = e[i]; 37 if(j == fa || st[j]) continue; 38 getrt(j, u); 39 size[u] += size[j]; 40 mx[u] = max(mx[u], size[j]); 41 } 42 mx[u] = max(mx[u], tot - size[u]); 43 if(mx[rt] > mx[u]) 44 rt = u; 45 } 46 47 void dfs(int u, int fa, int t) 48 { 49 t = 1ll * t * a[u] % mod; 50 tmp[++ top] = t; id[top] = u; 51 for (int i = h[u]; ~i; i = ne[i]) 52 { 53 int j = e[i]; 54 if(j == fa || st[j]) continue; 55 dfs(j, u, t); 56 } 57 } 58 59 void up(int x, int y) 60 { 61 x = 1ll * inv[x] * k % mod; 62 x = mp[x]; 63 if(x == 0) return ; 64 if(x > y) swap(x, y); 65 if(ans1 > x) 66 { 67 ans1 = x; 68 ans2 = y; 69 } 70 else if(ans1 == x && y < ans2) 71 { 72 ans2 = y; 73 } 74 } 75 76 void calc(int u, int fa) 77 { 78 mp[a[u]] = u; 79 for (int i = h[u]; ~i; i = ne[i]) 80 { 81 int j = e[i]; 82 if(j == fa || st[j]) continue; 83 top = 0; 84 dfs(j, u, 1); 85 for (int z = 1; z <= top; z ++) 86 up(tmp[z], id[z]); 87 top = 0; 88 dfs(j, u, a[u]); 89 for (int z = 1; z <= top; z ++) 90 { 91 int now = mp[tmp[z]]; 92 if(!now || id[z] < now) 93 mp[tmp[z]] = id[z]; 94 } 95 } 96 mp[a[u]] = 0; 97 for (int i = h[u]; ~i; i = ne[i]) 98 { 99 int j = e[i]; 100 if(j == fa || st[j]) continue; 101 top = 0; 102 dfs(j, u, a[u]); 103 for (int z = 1; z <= top; z ++) 104 mp[tmp[z]] = 0; 105 } 106 } 107 108 void solve(int u) 109 { 110 st[u] = 1; 111 calc(u, 0); 112 for (int i = h[u]; ~i; i = ne[i]) 113 { 114 int j = e[i]; 115 if(st[j]) continue; 116 rt = 0; 117 tot = size[j]; 118 getrt(j, 0); 119 solve(rt); 120 } 121 } 122 123 int main() 124 { 125 inv[1]=1; 126 for(int i=2;i< M;i++) 127 inv[i]= 1ll * (mod-mod/i)*inv[mod%i]%mod; 128 while(scanf("%d%d", &n, &k) != EOF) 129 { 130 ans1 = 1e9; ans2 = 1e9; 131 idx = 0; 132 for (int i = 1; i <= n; i ++) st[i] = 0, h[i] = -1; 133 for (int i = 1; i <= n; i ++) 134 a[i] = read(); 135 for (int i = 1; i <= n - 1; i ++) 136 { 137 int a = read(), b = read(); 138 add(a, b); add(b, a); 139 } 140 rt = 0; tot = mx[rt] = n; 141 getrt(1, 0); 142 solve(rt); 143 if(ans1 == 1e9) 144 { 145 puts("No solution"); 146 } 147 else 148 printf("%d %d ", ans1, ans2); 149 } 150 }