POJ1470.Closest Common Ancestors

题目链接：http://poj.org/problem?id=1470。
思路：给出一棵树和多个求两个节点的最近公共祖先的查询，使用tarjan 离线算法应对多对查询，使用并查集来合并顶点，不过这里不能够使用按秩合并，只能将root的所有孩子节点并到root上，代码如下：

#include <iostream>
#include <vector>
#include <cstring>
#include <cstdio>
using namespace std;

const int maxn = 905;
vector<int> graph[maxn], query[maxn];
int pre[maxn], visit[maxn], cnt[maxn], in[maxn];
int n, m;

void makeset(int root) {
    pre[root] = root;
}

int find(int root) {
    return (pre[root] == root) ? root : (pre[root] = find(pre[root]));
}

void Union(int s, int v) {
    int s1 = find(s);
    int v1 = find(v);

    pre[v1] = s1;
}

void tarjan(int root) {
    makeset(root);
    for (int i = 0; i < graph[root].size(); i++) {
        int v = graph[root][i];

        tarjan(v);
        Union(root, v);
    }
    visit[root] = 1;
    for (int i = 0; i < query[root].size(); i++) {
        int v = query[root][i];

        if (visit[v]) cnt[find(v)]++;
    }
}

int main() {
    int node, succ, num, s, e, root;
    while (cin >> n) {
    //init
        memset(pre, 0, sizeof(pre));
        memset(visit, 0, sizeof(visit));
        memset(cnt, 0, sizeof(cnt));
        memset(in, 0, sizeof(in));
        for (int i = 1; i <= n; i++) {
            graph[i].clear();
            query[i].clear();
        }
        for (int i = 0; i < n; i++) {
            scanf("%d:(%d)", &node, &num);
            for (int j = 0; j < num; j++) {
                cin >> succ;
                graph[node].push_back(succ);
                in[succ]++;
            }
        }
        cin >> m;
        for (int i = 0; i < m; i++) {
            while (getchar() != '(');
            cin >> s >> e;
            query[s].push_back(e);
            query[e].push_back(s);
            while (getchar() != ')');
        }
        for (int i = 1; i <= n; i++) {//the first input is not always the root node
            if (!in[i]) {
                root = i;
                break;
            }
        }
        tarjan(root);
        for (int i = 1; i <= n; i++) {
            if (cnt[i]) printf("%d:%d\n", i, cnt[i]);
        }
    }
    return 0;
}

参考链接：http://www.cnblogs.com/JVxie/p/4854719.html。