POJ 1470 Closest Common Ancestors (LCA)

题意：给定一棵树，然后有m个询问，最后统计公共祖先个数。

析：LCA，但是这个题输入太麻烦了，调试了好久，才出结果，然后就在Tarjan算法中直接统计就好了，刚开始MLE，后来又RE，没办法又换了一种方法才AC。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 9e2 + 5;
const int mod = 1e9 + 7;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}
vector<int> G[maxn];
vector<int> q[maxn];
int ans[maxn], p[maxn];
bool vis[maxn], in[maxn];

int Find(int x){  return x == p[x] ? x : p[x] = Find(p[x]); }

void Tarjan(int u){
    vis[u] = true;
    for(int i = 0; i < q[u].size(); ++i){
        int v = q[u][i];
        if(vis[v]) ++ans[Find(v)];
    }

    for(int i = 0; i < G[u].size(); ++i){
        int v = G[u][i];
        if(!vis[v]){
            Tarjan(v);
            p[v] = u;
        }
    }
}

int main(){
    while(scanf("%d", &n) == 1){
        for(int i = 1; i <= n;  ++i) G[i].clear(), q[i].clear(), p[i] = i;
        memset(in, false, sizeof in);
        int u, v;
        for(int i = 0; i < n; ++i){
            scanf("%d:(%d)", &u, &m);
            while(m--){
                scanf("%d", &v);
                in[v] = true;
                G[u].push_back(v);
                G[v].push_back(u);
            }
        }

        char s[5];
        scanf("%d", &m);
        while(m--){
            scanf("%1s%d %d%1s", s, &u, &v, s);
            q[u].push_back(v);
            q[v].push_back(u);
        }

        memset(ans, 0, sizeof ans);
        memset(vis, false, sizeof vis);
        for(int i = 1; i <= n; ++i)  if(!in[i]){
            Tarjan(i);
            break;
        }

        for(int i = 1; i <= n; ++i) if(ans[i])
            printf("%d:%d
", i, ans[i]);
    }
    return 0;
}