AtCoder Grand Contest 016 D

时间限制: 1 Sec 内存限制: 128 MB
提交: 3 解决: 2
[提交] [状态] [命题人:admin]

题目描述

There is a sequence of length N: a=(a1,a2,…,aN). Here, each ai is a non-negative integer.

Snuke can repeatedly perform the following operation:

Let the XOR of all the elements in a be x. Select an integer i (1≤i≤N) and replace ai with x.
Snuke's objective is to match a with another sequence b=(b1,b2,…,bN). Here, each bi is a non-negative integer.

Determine whether the objective is achievable, and find the minimum necessary number of operations if the answer is positive.

Constraints
2≤N≤105
ai and bi are integers.
0≤ai,bi<230

输入

Input is given from Standard Input in the following format:

N
a1 a2 … aN
b1 b2 … bN

输出

If the objective is achievable, print the minimum necessary number of operations. Otherwise, print -1 instead.

样例输入

3
0 1 2
3 1 0

样例输出

来源/分类

AtCoder Grand Contest 016

操作为选择一个数，把这个数的值换为整个序列的异或和，交换之后整个序列的异或和就变为这个数的值。

求出a和b中所有数的异或和,存在a[0],b[0]，这n+1个数如果不相同，那a就不可能与b相同。

如果两个序列中的数是相同的，那么当a[i]!=b[i]时，在a[i]与b[i]之间建一条边。这样最后就会得到一个图，把图中一个联通块的数换为与b相同花费的次数与联通块的大小相同，从一个联通块跳至另一个联通块需花费1。由于一开始我们可以移动的是a[0],如果a[0]不属于任何联通块，那么从a[0]跳至别的联通块需花费1。

#include "bits/stdc++.h"

using namespace std;
const int maxn = 1e5 + 100;
int a[maxn], b[maxn];
int vis[maxn];
map<int, int> mp;
vector<int> e[maxn];

void dfs(int now) {
    vis[now] = 1;
    for (auto p:e[now]) {
        if (!vis[p])
            dfs(p);
    }
}

int main() {
    freopen("input.txt", "r", stdin);
    int n;
    cin >> n;
    for (int i = 1; i <= n; i++) {
        cin >> a[i];
        a[0] ^= a[i];
    }
    for (int i = 1; i <= n; i++) {
        cin >> b[i];
        b[0] ^= b[i];
    }
    for (int i = 0; i <= n; i++) {
        mp[a[i]]++;
        mp[b[i]]--;
    }
    for (auto p:mp) {
        if (p.second != 0) {
            cout << -1 << endl;
            return 0;
        }
    }
    int id = 0;
    int ans = 0;
    for (int i = 0; i <= n; i++) {
        if (!mp[a[i]]) mp[a[i]] = ++id;
    }
    for (int i = 0; i <= n; i++) {
        if (a[i] != b[i]) {
            e[mp[a[i]]].push_back(mp[b[i]]), ans++;
        }
    }
    for (int i = 1; i <= id; i++) {
        if (!vis[i] && !e[i].empty()) {
            ans++;
            dfs(i);
        }
    }
    if (a[0] != b[0]) ans--;//a[0]和b[0]是不用花费一次交换使他们一样的，因为他们最终会变得一样
    if (!e[mp[a[0]]].empty()) ans--;
    cout << ans << endl;
    return 0;
}