【带权并查集】A Bug's Life POJ - 2492

A Bug's Life POJ - 2492

题意：

一种昆虫，异性之间才能配对，同性之间不能配对。现给出\(n\)只昆虫的\(m\)条配对关系，问其中有没有不合理（即同性配对）的情况。

思路：

虽然看起来迥异，但这题的内在逻辑和Parity game POJ - 1733是一样的……所以公式可以照搬，都是异或运算。具体推导表格可以直接看那一题的题解。

const int maxn = 200000 + 100;
int n, m;
int rela[maxn], fa[maxn];

int find(int x) {
    if (fa[x] == -1) return x;
    int tmp = find(fa[x]);
    rela[x] = rela[x] ^ rela[fa[x]];
    return fa[x] = tmp;
}

bool merge(int u, int v, int s) {
    int r1 = find(u);
    int r2 = find(v);
    if (r1 == r2) {
        return s == rela[u] ^ rela[v];
    }
    else {
        fa[r1] = r2;
        rela[r1] = rela[u] ^ rela[v] ^ s;
        return true;
    }
}

int main()
{
    //ios::sync_with_stdio(false);
    int t; cin >> t; for (int kase = 1; kase <= t; kase++) {
        n = read(); m = read();
        for (int i = 1; i <= n; i++) fa[i] = -1;
        memset(rela, 0, sizeof(rela));
        int ok = 1;
        for (int i = 1; i <= m; i++) {
            int u, v;
            u = read(); v = read();
            if (!ok) continue;
            if (!merge(u, v, 1)) ok = 0;
        }
        if (kase != 1) printf("\n");
        printf("Scenario #%d:\n", kase);
        if (ok) printf("No suspicious bugs found!\n");
        else printf("Suspicious bugs found!\n");
    }
    return 0;
}