方法1：用带边权的并查集

• 写法1

#include <iostream>
#include <unordered_map>

using namespace std;

const int N = 20010;

int n, m;
int p[N], d[N];
unordered_map<int, int> S;

int get(int x) {
    if (S.count(x) == 0) S[x] = ++n;
    return S[x];
}

int find(int x) {
    if (p[x] != x) {
        int root = find(p[x]);
        d[x] = d[x] ^ d[p[x]];
        p[x] = root;
    }
    return p[x];
}

int main() {

    cin >> n >> m;
    n = 0;
    for (int i = 0; i < N; i++) p[i] = i;

    int res = m;
    for (int i = 1; i <= m; i++) {
        int a, b;
        string type;
        cin >> a >> b >> type;
        a = get(a - 1), b = get(b);

        int t = 0;
        if (type == "odd") t = 1;

        int pa = find(a), pb = find(b);
        if (pa == pb) {  // 说明a和b在同一个集合中
            if ((d[a] ^ d[b]) != t) {
                res = i - 1;
                break;
            }
        } else {
            p[pa] = pb;
            d[pa] = d[a] ^ d[b] ^ t;
        }
    }

    cout << res << endl;

    return 0;
}

• 写法2

也可以采用下面的写法（d[x]存储x到父节点的距离，类似于食物链的做法）：

#include <iostream>
#include <unordered_map>

using namespace std;

const int N = 20010;

int n, m;
int p[N], d[N];
unordered_map<int, int> S;

int get(int x) {
    if (S.count(x) == 0) S[x] = ++n;
    return S[x];
}

int find(int x) {
    if (p[x] != x) {
        int root = find(p[x]);
        d[x] += d[p[x]];  // 可能会导致溢出
        p[x] = root;
    }
    return p[x];
}

int main() {

    cin >> n >> m;
    n = 0;
    for (int i = 0; i < N; i++) p[i] = i;

    int res = m;
    for (int i = 1; i <= m; i++) {
        int a, b;
        string type;
        cin >> a >> b >> type;
        a = get(a - 1), b = get(b);

        int t = 0;
        if (type == "odd") t = 1;

        int pa = find(a), pb = find(b);
        if (pa == pb) {  // 说明a和b在同一个集合中
            if (((d[a] + d[b]) % 2 + 2) % 2 != t) {  // 必须保证余数为负，也可以处理
                res = i - 1;
                break;
            }
        } else {
            p[pa] = pb;
            d[pa] = d[a] + d[b] + t;
        }
    }

    cout << res << endl;

    return 0;
}

方法2：带扩展域的并查集

#include <iostream>
#include <unordered_map>

using namespace std;

const int N = 40010, Base = N / 2;

int n, m;
int p[N], d[N];
unordered_map<int, int> S;

int get(int x) {
    if (S.count(x) == 0) S[x] = ++n;
    return S[x];
}

int find(int x) {
    if (p[x] != x) p[x] = find(p[x]);
    return p[x];
}

int main() {

    cin >> n >> m;
    n = 0;
    for (int i = 0; i < N; i++) p[i] = i;

    int res = m;
    for (int i = 1; i <= m; i++) {
        int a, b;
        string type;
        cin >> a >> b >> type;
        a = get(a - 1), b = get(b);

        if (type == "even") {
            if (find(a + Base) == find(b)) {
                // 此时题目输入为a,b是同类，但是发现a,b是异类，矛盾
                res = i - 1;
                break;
            }
            p[find(a)] = find(b);
            p[find(a + Base)] = find(b + Base);
        } else {
            if (find(a) == find(b)) {
                // 此时题目输入为a,b是异类，但是发现a,b是同类，矛盾
                res = i - 1;
                break;
            }
            p[find(a + Base)] = p[find(b)];
            p[find(a)] = p[find(b + Base)];
        }
    }

    cout << res << endl;

    return 0;
}