标签:style blog io color ar os sp for div
题意给一颗树,再给一个查询两点之间的最近公共祖先。
#include<iostream> #include<cstdio> #include<cstring> #include<map> #include<vector> using namespace std; const int maxn = 111111; struct edge { int to; int next; }e[maxn * 10]; int len; int head[maxn]; struct Node { int val; int id; }vis[maxn], dp[maxn][20]; int pos[maxn]; int cnt; void add(int from, int to) { e[len].to = to; e[len].next = head[from]; head[from] = len++; } void dfs(int x, int val) { vis[cnt].val = val; vis[cnt].id = x; pos[x] = cnt++; for (int i = head[x]; i != -1; i = e[i].next){ int cc = e[i].to; dfs(cc, val + 1); vis[cnt].val = val; vis[cnt].id = x; pos[x] = cnt++; } } void init(int k) { for (int i = 0; i < k; i++) dp[i][0] = vis[i]; for (int j = 1; (1 << j) <= k; j++){ for (int i = 0; i + (1 << j) - 1 < k; i++){ if (dp[i][j - 1].val < dp[i + (1 << (j - 1))][j - 1].val) dp[i][j] = dp[i][j - 1]; else dp[i][j] = dp[i + (1 << (j - 1))][j - 1]; } } } int ask(int l, int r) { int k = 0; while ((1 << (k + 1)) < r - l + 1) k++; if (dp[l][k].val < dp[r - (1 << k) + 1][k].val) return dp[l][k].id; else return dp[r - (1 << k) + 1][k].id; } int main() { int T; int n; cin >> T; int root; int gg[maxn]; int a, b; while (T--){ cin >> n; len = 0; memset(head, -1, sizeof(head)); for (int i = 1; i <= n; i++) gg[i] = 0; for (int i = 0; i<n - 1; i++){ scanf("%d%d", &a, &b); add(a, b); gg[b] = 1; } for (int i = 1; i <= n; i++) if (!gg[i]) { root = i; break; } cnt = 0; dfs(root, 1); init(cnt); cin >> a >> b; int l = pos[a]; int r = pos[b]; if (l>r) swap(l, r); cout << ask(l, r) << endl; } return 0; }
Poj1330Nearest Common Ancestors LCA
标签:style blog io color ar os sp for div
原文地址:http://www.cnblogs.com/yigexigua/p/4083566.html