uva12489 Combating cancer(树同构)

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https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=3933

给你两棵无根树,让你判断这两棵树是否同构

不会判断树同构,果断抄了个模板,乱搞给过掉了。

首先由于给的是无根树,而要判断无根树是否同构得以重心为根,然后做一个括号序列的哈希。

于是我们需要先找出重心,要找树的重心得先知道直径。

找出直径,直径上的点的个数是偶数,那么重心是中间的两个点,如果是奇数个,那么重心是中间那个。

或者说是根据拓排,每次度数为1的点入队,留下的最后一批就是。

然而我当时脑抽了一下,求好直径,后用第二种再去搞。。。其实求直径的时候保存一下路径就好了。

最后,如果有两个重心就做两次哈希,得到两个哈希值,一个就一次。

最后把两棵树的哈希值比一下是否有相同的。

/**
 * code generated by JHelper
 * More info: https://github.com/AlexeyDmitriev/JHelper
 * @author xyiyy @https://github.com/xyiyy
 */

#include <iostream>
#include <fstream>

//#####################
//Author:fraud
//Blog: http://www.cnblogs.com/fraud/
//#####################
//#pragma comment(linker, "/STACK:102400000,102400000")
#include <iostream>
#include <sstream>
#include <ios>
#include <iomanip>
#include <functional>
#include <algorithm>
#include <vector>
#include <string>
#include <list>
#include <queue>
#include <deque>
#include <stack>
#include <set>
#include <map>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <climits>
#include <cctype>

using namespace std;
#define pb(X) push_back(X)
#define rep(X, N) for(int X=0;X<N;X++)
#define ALL(X) (X).begin(),(X).end()
typedef unsigned long long ull;

const int maxNode = 10010;
const int maxEdge = (maxNode << 1);
//
// Created by xyiyy on 2015/8/14.
//

#ifndef ICPC_HASHTREE_HPP
#define ICPC_HASHTREE_HPP

//树的同构,返回哈希值
//输入有根树的根,或者无根树的重心
typedef unsigned long long ull;
const ull MAGIC = 321;

//
// Created by xyfra_000 on 2015/8/14.
//

#ifndef ICPC_ADJLIST_ARRAY_HPP
#define ICPC_ADJLIST_ARRAY_HPP

#define Foredge(A, X) for(int A = From[X];A!=-1;A = Next[A])

int From[maxEdge], To[maxEdge];
int Next[maxEdge];
int Edgecnt;

void init(int n) {
    rep(i, n + 1)From[i] = -1;
    Edgecnt = 0;
}

void addedge(int u, int v) {
    To[Edgecnt] = v;
    Next[Edgecnt] = From[u];
    From[u] = Edgecnt++;
}

#endif //ICPC_ADJLIST_ARRAY_HPP


ull powMod(ull a, int n) {
    ull ret = 1ULL;
    while (n) {
        if (n & 1)ret *= a;
        a *= a;
        n >>= 1;
    }
    return ret;
}

struct Hash {
    int length;
    ull value;

    Hash() : length(0), value(0) { }

    Hash(char c) : length(1), value(c) { }

    Hash(int l, int v) : length(l), value(v) { }
};

bool operator<(const Hash &a, const Hash &b) {
    return a.value < b.value;
}

Hash operator+(const Hash &a, const Hash &b) {
    return Hash(a.length + b.length, a.value * powMod(MAGIC, b.length) + b.value);
}

void operator+=(Hash &a, Hash &b) {
    a = a + b;
}

vector<Hash> childs[maxNode];

Hash dfs(int pre, int cur) {
    Hash ret;
    childs[cur].clear();
    for (int iter = From[cur]; iter != -1; iter = Next[iter]) {
        if (To[iter] != pre) {
            childs[cur].pb(dfs(cur, To[iter]));
        }
    }
    sort(ALL(childs[cur]));
    for (vector<Hash>::iterator iter = childs[cur].begin(); iter != childs[cur].end(); iter++) {
        ret += *iter;
    }
    Hash retL = Hash('(');
    ret = '(' + ret + ')';
    return ret;
}

ull getHash(int root) {
    return dfs(-1, root).value;
}

#endif //ICPC_HASHTREE_HPP

//
// Created by xyfra_000 on 2015/8/14.
//

#ifndef ICPC_TREEDIAMETER_HPP
#define ICPC_TREEDIAMETER_HPP

//求树的直径
//可以通过修改dfs部分变成求带权的树的直径

vector<int> dist;

void dfs(int p, int u, int d) {
    dist[u] = d;
    Foredge(i, u) {
        if (To[i] != p) {
            dfs(u, To[i], d + 1);
        }
    }
}

int getDiameter(int n) {
    dist.resize(n);
    dfs(-1, 0, 0);
    int u = max_element(ALL(dist)) - dist.begin();
    dfs(-1, u, 0);
    return *max_element(ALL(dist));
}


#endif //ICPC_TREEDIAMETER_HPP

int deg[maxNode];
int vis[maxNode];

class TaskH {
public:
    void solve(std::istream &in, std::ostream &out) {
        int n;
        while (in >> n) {
            vector<ull> ans[2];
            rep(times, 2) {
                int u, v;
                init(n);
                rep(i, n + 1)deg[i] = 0;
                rep(i, n + 1)vis[i] = 0;
                queue<int> q;
                rep(i, n - 1) {
                    in >> u >> v;
                    u--, v--;
                    deg[u]++;
                    deg[v]++;
                    addedge(u, v);
                    addedge(v, u);
                }
                int dia = getDiameter(n);
                int num = n;
                rep(i, n) {
                    if (deg[i] == 1) {
                        q.push(i);
                    }
                }
                int gao = 1;
                if (dia & 1)gao++;
                while (num > gao && !q.empty()) {
                    u = q.front();
                    q.pop();
                    vis[u] = 1;
                    num--;
                    deg[u]--;
                    for (int i = From[u]; i != -1; i = Next[i]) {
                        int v = To[i];
                        if (!vis[v]) {
                            deg[v]--;
                            if (deg[v] == 1) {
                                q.push(v);
                            }
                        }
                    }
                }
                rep(i, n) {
                    if (!vis[i]) {
                        ans[times].pb(getHash(i));
                    }
                }
            }
            bool ok = 0;
            rep(i, ans[0].size()) {
                rep(j, ans[1].size()) {
                    if (ans[0][i] == ans[1][j])ok = 1;
                }
            }
            if (ok)out << "S" << endl;
            else out << "N" << endl;
        }
    }
};

int main() {
    std::ios::sync_with_stdio(false);
    std::cin.tie(0);
    TaskH solver;
    std::istream &in(std::cin);
    std::ostream &out(std::cout);
    solver.solve(in, out);
    return 0;
}

 

posted on 2015-08-14 20:46  xyiyy  阅读(...)  评论(...编辑  收藏

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