HDU 6625 three arrays or minimum sum of two sequences exclusive arrangement (a replicable Positive Solutions

@ (Hdu 6625 the sum of two aligned sequences XOR minimum)

Meaning of the questions:

\ (T (100) \) group, two of each length \ (n (100000) \) arrangement, you can \ (A [] \) and \ (B [] \) are randomly arranged, can be obtained \ (C [I] = a [I] \) ^ \ (B [I] \) , find lexicographically smallest \ (C [] \) .

Resolve

A greedy approach is clearly right:
for this question

• Every two trie to go down at the same time, if there \ (0 \) or \ (1 \) this path, while casually walking \ (0 \; or \; 1 \) this path, otherwise only can go a \ (0 \) , a walk \ (1 \) . Such complexity is strictly \ (O (log) \) , and finally the resulting \ (n \) digital sort is the final answer.
• Why is it so right?
• If the current tree has two dictionaries \ (0 \) and \ (1 \) path, while walking \ (0 \) this road is certainly not guaranteed to get numbers can be different or out of the current minimum, but be sure he is certainly one of the values lexicographically smallest sequence contains.
• If you want to ask two simple 01 trie XOR minimum personal feeling is not good practice complexity.

It can promote a positive solution:

• The topic and people \ (dreamoon \) positive solutions provide:
• Now \ (A [] \) in stuffing a digital \ (X \) , then \ (B [] \) corresponding to find a and \ (X \) matching digits XOR minimum \ (Y \) , then in the \ (a [] \) to find an inside and \ (Y \) matching the lowest number \ (Z \) , recursion will find a ring of size 2 .
• This ring these two figures taken out, and then back to a mismatch position to continue recursion.
• Thus obtained \ (n \) digital sort after for the final answer.
• Complexity is also very scientific and very broad applicability of this idea.

Code1

const int MXN = 1e5 + 7;
const int MXE = 2e6 + 7;
int n, m;
int ar[MXN], br[MXN];
struct Trie {
    int tot;
    int nex[MXE][2], num[MXE], val[MXE];
    Trie(){nex[0][0] = nex[0][1] = -1;}
    void newnode() {
        ++ tot;
        nex[tot][0] = nex[tot][1] = -1;
    }
    void inisert(int x) {
        int rt = 0;
        for(int i = 31, tmp; i >= 0; --i) {
            tmp = ((x>>i)&1);
            if(nex[rt][tmp] == -1) newnode(), nex[rt][tmp] = tot;
            rt = nex[rt][tmp];
            num[rt] ++;
        }
        val[rt] = x;
    }
    void del(int x) {
        int rt = 0;
        for(int i = 31, tmp; i >= 0; --i) {
            tmp = ((x>>i)&1);
            int lst = rt;
            rt = nex[rt][tmp];
            nex[lst][tmp] = -1;
            num[rt] = 0;
        }
    }
}cw[2];
bool check(int id, int rt, int tmp) {
    return cw[id].nex[rt][tmp] != -1 && cw[id].num[cw[id].nex[rt][tmp]] > 0;
}
int getans() {
    int rt1 = 0, rt2 = 0;
    for(int i = 31; i >= 0; --i) {
        if(check(0, rt1, 0) && check(1, rt2, 0)) {
            rt1 = cw[0].nex[rt1][0];
            rt2 = cw[1].nex[rt2][0];
            -- cw[0].num[rt1];
            -- cw[1].num[rt2];
        }else if(check(0, rt1, 1) && check(1, rt2, 1)) {
            rt1 = cw[0].nex[rt1][1];
            rt2 = cw[1].nex[rt2][1];
            -- cw[0].num[rt1];
            -- cw[1].num[rt2];
        }else if(check(0, rt1, 1) && check(1, rt2, 0)) {
            rt1 = cw[0].nex[rt1][1];
            rt2 = cw[1].nex[rt2][0];
            -- cw[0].num[rt1];
            -- cw[1].num[rt2];
        }else if(check(0, rt1, 0) && check(1, rt2, 1)) {
            rt1 = cw[0].nex[rt1][0];
            rt2 = cw[1].nex[rt2][1];
            -- cw[0].num[rt1];
            -- cw[1].num[rt2];
        }
    }
    return cw[0].val[rt1] ^ cw[1].val[rt2];
}
int main() {
#ifndef ONLINE_JUDGE
    freopen("/home/cwolf9/CLionProjects/ccc/in.txt", "r", stdin);
//    freopen("/home/cwolf9/CLionProjects/ccc/out.txt", "w", stdout);
#endif
    int tim = read();
    while(tim --) {
        n = read();
        cw[0].tot = cw[1].tot = 0;
        for(int i = 1; i <= n; ++i) ar[i] = read(), cw[0].inisert(ar[i]);
        for(int i = 1; i <= n; ++i) br[i] = read(), cw[1].inisert(br[i]);
        vector<int> vs;
        for(int i = 1; i <= n; ++i) vs.eb(getans());
        sort(all(vs));
        for(int i = 0; i < SZ(vs); ++i) printf("%d%c", vs[i], " \n"[i == SZ(vs) - 1]);
        for(int i = 1; i <= n; ++i) cw[0].del(ar[i]), cw[1].del(br[i]);
    }
    return 0;
}

Code2

const int MXN = 1e5 + 7;
const int MXE = 2e6 + 7;
int n, m;
int ar[MXN], br[MXN];
struct Trie {
    int tot;
    int nex[MXE][2], num[MXE], val[MXE];
    Trie(){nex[0][0] = nex[0][1] = -1;}
    void newnode() {
        ++ tot;
        nex[tot][0] = nex[tot][1] = -1;
    }
    void inisert(int x) {
        int rt = 0;
        for(int i = 30, tmp; i >= 0; --i) {
            tmp = ((x>>i)&1);
            if(nex[rt][tmp] == -1) newnode(), nex[rt][tmp] = tot;
            rt = nex[rt][tmp];
            num[rt] ++;
        }
        val[rt] = x;
    }
    int query(int x) {
        int rt = 0;
        for(int i = 30, tmp; i >= 0; --i) {
            tmp = ((x>>i)&1);
            if(nex[rt][tmp] != -1 && num[nex[rt][tmp]]) rt = nex[rt][tmp];
            else rt = nex[rt][!tmp];
        }
        return val[rt];
    }
    int find() {
        int rt = 0;
        for(int i = 30, tmp; i >= 0; --i) {
            if(nex[rt][0] != -1 && num[nex[rt][0]]) rt = nex[rt][0];
            else if(nex[rt][1] != -1 && num[nex[rt][1]]) rt = nex[rt][1];
        }
        if(rt == 0) return -1;
        return val[rt];
    }
    void del() {
        for(int i = 0; i <= tot + 1; ++i) num[i] = 0, clr(nex[i], -1);
        tot = 0;
    }
    void sub(int x) {
        int rt = 0;
        for(int i = 30, tmp; i >= 0; --i) {
            tmp = ((x>>i)&1);
            rt = nex[rt][tmp];
            num[rt] --;
        }
    }
}cw[2];
/*
 * 这种做法不能保证每次求出来的异或最小值都是单调递增的，但是将n次得到的值排序后一定是正确答案
 * 如果想单纯的求两个01字典树异或最小值，个人感觉还没有较好的复杂度的做法。
 * 关于本题，还有一个出题人提供适用性更加广泛的正解：
 * 现在a中随便找一个数字，然后在b中找一个和他匹配最小的数字，再在a里面找一个和上个数匹配最小的数字，递归下去一定会找到一个大小为2的环
 * 把这个环取出来，在回到上一个位置继续递归下去。得到的n个数字排序即为最终答案。
 * */
vector<int> vs;
int dfs(int id, int x, int lst) {
    int tmp = cw[!id].query(x);
    if(tmp == lst) {
        vs.eb(tmp ^ x);
        cw[id].sub(x);
        cw[!id].sub(tmp);
        return id;
    }
    int ret = dfs(!id, tmp, x);
    if(ret != id) return ret;
}
int main() {
#ifndef ONLINE_JUDGE
    freopen("/home/cwolf9/CLionProjects/ccc/in.txt", "r", stdin);
//    freopen("/home/cwolf9/CLionProjects/ccc/out.txt", "w", stdout);
#endif
    int tim = read();
    while(tim --) {
        n = read();
        for(int i = 1; i <= n; ++i) ar[i] = read(), cw[0].inisert(ar[i]);
        for(int i = 1; i <= n; ++i) br[i] = read(), cw[1].inisert(br[i]);
        vs.clear();
        while(1) {
            int tmp = cw[0].find();
            if(tmp == -1) break;
            dfs(1, tmp, -1);
        }
        sort(all(vs));
        for(int i = 0; i < SZ(vs); ++i) printf("%d%c", vs[i], " \n"[i == SZ(vs) - 1]);
        cw[0].del(), cw[1].del();
    }
    return 0;
}

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