AGC018F. Two Trees

题意

给出两棵树，要求给两棵树上相同编号的点赋值，使得每个点的子树权值和的绝对值为１

做法

如果某个编号代表的点在一棵树中的儿子数为奇数而在另一棵树中的儿子数为偶数，那么无论这个位置填什么都不可能，否则可以构造只填０，１，－１的方案。具体做法是，分别考虑两棵树，如果一个点的儿子数为奇数，那么这个点填０；否则对于这个点所在的子树里的奇数儿子点两两配对，表示一个填１另一个填－１。把两棵树的配对情况一起建图，最后得到的必然是二分图，因为任何一个环上的边必然是交替来自两棵树内的配对。故而二染色填数即可。

#include <bits/stdc++.h>
using namespace std;

const int N = 1e5 + 100;

int n, fa[N], fb[N], is_odd[2][N];

struct Edge {
    int e, *h, *nx, *to, V, E;
    Edge(int V, int E):V(V), E(E) {
        e = 0;
        h = new int [V];
        nx = new int [E];
        to = new int [E];
        memset(h, -1, 4 * V);
    }
    ~Edge() {
        delete h;
        delete nx;
        delete to;
    }
    void clear() {
        e = 0;
        memset(h, -1, 4 * V);
    }
    void add(int u, int v) {
        to[e] = v, nx[e] = h[u], h[u] = e++;
        to[e] = u, nx[e] = h[v], h[v] = e++;
    }
} *tr, *bi;

int dfs(int u, int f) {
    vector<int> odd_son;
    for (int i = tr->h[u]; i != -1; i = tr->nx[i]) {
        int v = tr->to[i];
        if (v != f) {
            v = dfs(v, u);
            if (v != 0)
                odd_son.push_back(v);
        }
    }
    if (!is_odd[0][u])
        odd_son.push_back(u);
    for (int i = 0; i < (int)odd_son.size() - 1; i += 2) {
        bi->add(odd_son[i], odd_son[i + 1]);
    }
    return odd_son.size() % 2 == 0 ? 0 : odd_son.back();
}

bool vis[N], col[N];

void coloring(int u, int c) {
    vis[u] = true;
    col[u] = c;
    for (int i = bi->h[u]; i != -1; i = bi->nx[i]) {
        int v = bi->to[i];
        if (!vis[v])
            coloring(v, c ^ 1);
    }
}

int main() {
#ifdef lol
    freopen("f.in", "r", stdin);
    freopen("f.out", "w", stdout);
#endif

    scanf("%d", &n);
    for (int i = 1; i <= n; ++i) {
        scanf("%d", &fa[i]);
        if (fa[i] != -1) {
            is_odd[0][fa[i]] ^= 1;
        }
    }
    for (int i = 1; i <= n; ++i) {
        scanf("%d", &fb[i]);
        if (fb[i] != -1)
            is_odd[1][fb[i]] ^= 1;
    }
    for (int i = 1; i <= n; ++i) {
        if (is_odd[0][i] ^ is_odd[1][i]) {
            puts("IMPOSSIBLE");
            return 0;
        }
    }
    tr = new Edge(n + 1, (n + 1) * 2);
    bi = new Edge(n + 10, (n + 10) * 4);
    int rt;
    for (int i = 1; i <= n; ++i) {
        if (fa[i] == -1) {
            rt = i;
            continue;
        }
        tr->add(i, fa[i]);
    }
    dfs(rt, -1);
    tr->clear();
    for (int i = 1; i <= n; ++i) {
        if (fb[i] == -1) {
            rt = i;
            continue;
        }
        tr->add(i, fb[i]);
    }
    dfs(rt, -1);
    for (int i = 1; i <= n; ++i)
        if (!vis[i]) {
            coloring(i, 1);
        }
    puts("POSSIBLE");
    for (int i = 1; i <= n; ++i) {
        if (is_odd[0][i]) {
            printf("%d ", 0);
        } else {
            printf("%d ", col[i] ? 1 : -1);
        }
    }
    puts("");

    return 0;
}