http://acm.hdu.edu.cn/showproblem.php?pid=3078
题意：给出n个点n-1条边m个询问，每个点有个权值，询问中有k，u，v，当k = 0的情况是将u的权值修改成v，当k不为0的情况是问u和v的路径中权值第k大的点的权值是多少。
思路：比较暴力的方法，可能数据太水勉强混过去了。对于每一个询问的时候保留两个点之间的lca，还有计算出两个点之间的点的个数（询问的时候如果点的个数小于k就不用算了），然后tarjan算完之后对每个询问再暴力路径上的每个点放进vector排序输出第k大..
1 #include <cstdio> 2 #include <cstring> 3 #include <algorithm> 4 #include <iostream> 5 #include <queue> 6 #include <cmath> 7 #include <map> 8 #include <vector> 9 using namespace std; 10 #define N 80010 11 #define M 30010 12 struct node 13 { 14 int u, v, next; 15 }edge[N*2]; 16 struct P 17 { 18 int u, v, next, num, k, lca; 19 }Edge[M*2]; 20 int n, m, tot, Tot, head[N], Head[N], a[N], fa[N], dis[N]; 21 bool vis[N]; 22 vector<int> vec; 23 24 void init() 25 { 26 tot = Tot = 0; 27 memset(head, -1, sizeof(head)); 28 memset(Head, -1, sizeof(Head)); 29 memset(dis, -1, sizeof(dis)); 30 memset(vis, false, sizeof(vis)); 31 for(int i = 1; i <= n; i++) fa[i] = i; 32 } 33 34 void add(int u, int v) 35 { 36 edge[tot].u = u; edge[tot].v = v; edge[tot].next = head[u]; head[u] = tot++; 37 } 38 39 void Add(int u, int v, int k) 40 { 41 Edge[Tot].k = k; Edge[Tot].u = u; Edge[Tot].v = v; Edge[Tot].next = Head[u]; Edge[Tot].num = 0; Edge[Tot].lca = -1; Head[u] = Tot++; 42 } 43 44 int Find(int x) 45 { 46 while(x != fa[x]) x = fa[x]; 47 return x; 48 } 49 50 void Merge(int u, int v) 51 { 52 u = Find(u), v = Find(v); 53 fa[v] = u; 54 } 55 56 void tarjan(int u) 57 { 58 vis[u] = true; 59 for(int i = head[u]; ~i; i = edge[i].next) { 60 int v = edge[i].v; 61 if(!vis[v]) { 62 dis[v] = dis[u] + 1; 63 tarjan(v); 64 fa[v] = u; 65 } 66 } 67 for(int i = Head[u]; ~i; i = Edge[i].next) { 68 int v = Edge[i].v, k = Edge[i].k; 69 if(vis[v]) { 70 int ff = Find(v); 71 if(k) { 72 Edge[i].num = Edge[i^1].num = dis[u] + dis[v] - 2 * dis[ff]; 73 Edge[i].lca = Edge[i^1].lca = ff; 74 } 75 else a[u] = v; 76 } 77 } 78 } 79 80 int doit(int u, int v, int k, int num, int lca) 81 { 82 vec.clear(); 83 while(u != lca) { 84 vec.push_back(a[u]); u = fa[u]; 85 } 86 while(v != lca) { 87 vec.push_back(a[v]); v = fa[v]; 88 } 89 vec.push_back(a[lca]); 90 sort(vec.begin(), vec.end()); 91 num++; 92 return vec[num-k]; 93 } 94 95 void solve() 96 { 97 for(int i = 0; i < Tot; i += 2) { 98 int u = Edge[i].u, v = Edge[i].v, k = Edge[i].k, num = Edge[i].num, lca = Edge[i].lca; 99 if(k) { 100 if(num + 1 < k) puts("invalid request!"); 101 else printf("%d ", doit(u, v, k, num, lca)); 102 } 103 } 104 } 105 106 int main() 107 { 108 scanf("%d%d", &n, &m); 109 init(); 110 for(int i = 1; i <= n; i++) scanf("%d", a+i); 111 for(int i = 1; i < n; i++) { 112 int u, v; 113 scanf("%d%d", &u, &v); 114 add(u, v); add(v, u); 115 } 116 for(int i = 0; i < m; i++) { 117 int u, v, k; 118 scanf("%d%d%d", &k, &u, &v); 119 Add(u, v, k); Add(v, u, k); 120 } 121 dis[1] = 0; 122 tarjan(1); 123 solve(); 124 return 0; 125 }