题干：
You are given a tree (an undirected acyclic connected graph) with N nodes, and edges numbered 1, 2, 3...N-1. Each edge has an integer value assigned to it, representing its length.
We will ask you to perfrom some instructions of the following form:
DIST a b : ask for the distance between node a and node b orKTH a b k : ask for the k-th node on the path from node a to node bExample:N = 6 1 2 1 // edge connects node 1 and node 2 has cost 1 2 4 1 2 5 2 1 3 1 3 6 2 Path from node 4 to node 6 is 4 -> 2 -> 1 -> 3 -> 6 DIST 4 6 : answer is 5 (1 + 1 + 1 + 2 = 5) KTH 4 6 4 : answer is 3 (the 4-th node on the path from node 4 to node 6 is 3)
InputThe first line of input contains an integer t, the number of test cases (t <= 25). ttest cases follow.
For each test case:
In the first line there is an integer N (N <= 10000)In the next N-1 lines, the i-th line describes the i-th edge: a line with three integers a b c denotes an edge between a, b of cost c (c <= 100000)The next lines contain instructions "DIST a b" or "KTH a b k"The end of each test case is signified by the string "DONE".There is one blank line between successive tests.
OutputFor each "DIST" or "KTH" operation, write one integer representing its result.
Print one blank line after each test.
Example Input:161 2 12 4 12 5 21 3 13 6 2DIST 4 6KTH 4 6 4DONEOutput:53解题报告：
AC代码：
#include<cstdio>#include<iostream>#include<algorithm>#include<queue>#include<map>#include<queue>#include<stack>#include<vector>#include<set>#include<string>#include<cmath>#include<cstring>#define ll long long#define pb push_back#define pm make_pair#define fi first#define se secondusing namespace std;const int MAX = 20000 + 5;int dep[MAX],fa[MAX][33],cost[MAX][33];int n,m;vector<int> vv[MAX];vector<int> ww[MAX];void dfs(int cur, int rt) {fa[cur][0] = rt;dep[cur] = dep[rt] + 1;for(int i = 1; i < 31; ++i) {fa[cur][i] = fa[fa[cur][i - 1]][i - 1];cost[cur][i] = cost[fa[cur][i - 1]][i - 1] + cost[cur][i - 1];}int sz = vv[cur].size();for (int i = 0; i < sz; ++i) {if (vv[cur][i] == rt) continue;cost[vv[cur][i]][0] = ww[cur][i];dfs(vv[cur][i], cur);}}int lca(int u,int v) {if(dep[u] < dep[v]) swap(u,v);int res = 0;int dc = dep[u]-dep[v];for(int i = 0; i<=30; i++) {if((1<<i) & dc) res += cost[u][i],u=fa[u][i];}if(u == v) return res;for(int i = 30; i>=0 && u!=v; i--) {if(fa[u][i] != fa[v][i]) {res += cost[v][i] + cost[u][i];u=fa[u][i];v=fa[v][i];}}res += cost[u][0] + cost[v][0];u=fa[u][0];return res;}int fk(int a,int b,int k) {int u=a,v=b;if(dep[u] < dep[v]) swap(u,v);int dc = dep[u]-dep[v];for(int i = 0; i<=30; i++) {if((1<<i) & dc) u=fa[u][i];}if(u!=v) {for(int i = 30; i>=0 && u!=v; i--) {if(fa[u][i] != fa[v][i]) {u=fa[u][i];v=fa[v][i];}}u=fa[u][0];//得到公共祖先}int cur = 0;int ans = 0;if(dep[a] - dep[u] >= k) {int RES = k-1;if(RES!=0) {for(int i = 30; i>=0; i--) {int now = fa[a][i];if(dep[a] - dep[now] < RES) {int tmp = dep[a] - dep[now];a = fa[a][i];RES -= tmp;}}ans = fa[a][0];} else ans = a; } else if(dep[a] - dep[u] == k-1) {ans = u;} else {int RES = k - (dep[a] - dep[u])-1;RES = (dep[b]-dep[u]-RES);if(RES != 0) {for(int i = 30; i>=0; i--) {int now = fa[b][i];if(dep[b] - dep[now] < RES) {int tmp = dep[b] - dep[now];b = fa[b][i];RES -= tmp;}}ans = fa[b][0];}else ans = b;}return ans;}int main() {int t;cin>>t;while(t--) {scanf("%d",&n);memset(dep,0,sizeof dep);memset(fa,0,sizeof fa);memset(cost,0,sizeof cost);for(int i = 1; i<=n; i++) vv[i].clear(),ww[i].clear();for(int a,b,c,i = 1; i<=n-1; i++) {scanf("%d%d%d",&a,&b,&c);vv[a].pb(b);vv[b].pb(a);ww[a].pb(c);ww[b].pb(c);}dfs(1,-1);char op[22];while(scanf("%s",op)) {int a,b,k;if(!strcmp(op,"DIST")) {scanf("%d%d",&a,&b);printf("%d\n",lca(a,b));} else if(!strcmp(op,"KTH")) {scanf("%d%d%d",&a,&b,&k);printf("%d\n",fk(a,b,k));} else break;}if(t) puts("");}return 0 ;}
