SPOJ QTREE2Query on a tree II（LCA）

题意:

给定N≤104的一棵树,现有一些查询
DIST a b 询问(a,b)的距离
KTH a b k 询问(a,b)路径上的第k个点

分析:

DIST询问LCA直接搞,第二个倍增一下搞就好了

代码:

////  Created by TaoSama on 2015-11-09//  Copyright (c) 2015 TaoSama. All rights reserved.////#pragma comment(linker, "/STACK:1024000000,1024000000")#include <algorithm>#include <cctype>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iomanip>#include <iostream>#include <map>#include <queue>#include <string>#include <set>#include <vector>using namespace std;#define pr(x) cout << #x << " = " << x << "  "#define prln(x) cout << #x << " = " << x << endlconst int N = 1e4 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;int n;const int LOG = 15;int p[LOG][N], dep[N], dp[N];struct Edge {    int v, c, nxt;} edge[N << 1];int head[N], cnt;void add_edge(int u, int v, int c) {    edge[cnt] = (Edge) {v, c, head[u]};    head[u] = cnt++;}void dfs(int u, int f, int d) {    dep[u] = d;    p[0][u] = f;    for(int i = head[u]; ~i; i = edge[i].nxt) {        int v = edge[i].v;        if(v == f) continue;        dp[v] = dp[u] + edge[i].c;        dfs(v, u, d + 1);    }}void build() {    dfs(1, -1, 0);    for(int i = 0; i + 1 < LOG; ++i)        for(int v = 1; v <= n; ++v)            if(p[i][v] < 0) p[i + 1][v] = -1;            else p[i + 1][v] = p[i][p[i][v]];}int lca(int u, int v) {    if(dep[u] > dep[v]) swap(u, v);    for(int i = 0; i < LOG; ++i)        if(dep[v] - dep[u] >> i & 1)            v = p[i][v];    if(u == v) return u;    for(int i = LOG - 1; ~i; --i) {        if(p[i][u] != p[i][v]) {            u = p[i][u];            v = p[i][v];        }    }    return p[0][u];}int get(int u, int k) {    for(int i = 0; i < LOG; ++i)        if(k >> i & 1) u = p[i][u], k ^= 1 << i;    return u;}int main() {#ifdef LOCAL    freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);//  freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);#endif    ios_base::sync_with_stdio(0);    int t; scanf("%d", &t);    while(t--) {        scanf("%d", &n);        cnt = 0; memset(head, -1, sizeof head);        for(int i = 1; i < n; ++i) {            int u, v, c; scanf("%d%d%d", &u, &v, &c);            add_edge(u, v, c);            add_edge(v, u, c);        }        build();        while(true) {            char op[5]; scanf("%s", op);            if(op[1] == 'O') break;            int u, v; scanf("%d%d", &u, &v);            if(op[1] == 'I') {                int _lca = lca(u, v);                printf("%d\n", dp[u] + dp[v] - 2 * dp[_lca]);            } else {                int k; scanf("%d", &k);                --k;                int _lca = lca(u, v);                int l = dep[u] - dep[_lca];                int d = dep[u] + dep[v] - 2 * dep[_lca];                if(k == 0) printf("%d\n", u);                else if(k == l) printf("%d\n", _lca);                else if(k < l) printf("%d\n", get(u, k));                else if(k == d) printf("%d\n", v);                else {                    k = d - k;                    printf("%d\n", get(v, k));                }            }        }    }    return 0;}