zoukankan      html  css  js  c++  java
  • Poj1330Nearest Common Ancestors LCA

    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;
    }
  • 相关阅读:
    判断UpLoader是否安装了Flash
    事务
    AMQP
    分布式领域CAP理论
    查看数据库所有表的所有字段
    拼分页方法
    Website English Comments
    SQL语句执行时间测试
    一般处理程序返回json
    MVC Action返回Json
  • 原文地址:https://www.cnblogs.com/yigexigua/p/4083566.html
Copyright © 2011-2022 走看看