洛谷 P5937 [CEOI1999] Parity Game

题意：区间长度为n，m个查询。每次查询给出区间与一个数值0或者1，代表区间内的1的个数。找出不矛盾的最后一个询问。

思路：首先用到区间压缩，排序后去重即可。使用带权dsu，如果是同一个root，那么xor运算看是否符合输入。如果不是同一个root，直接合并。这里合并区间的时候权重更新有点抽象，xx合并到yy，使用的是x->xx的权重和y->yy的权重以及设定的op。相较于路径权重是长度时，稍微难理解一点。

class DisjointSet{
public:
    DisjointSet(int sz) : sz_(sz){
        fa_.resize(sz_);
        iota(fa_.begin(), fa_.end(), 0);
        dist_.resize(sz_);
    }

    int findSet(int x){
        updateDist(x);
        return fa_[x];
    }

    bool unionSet(int x, int y, int value){
        int xx = findSet(x);
        int yy = findSet(y);
        if (xx == yy){ return false; }
        fa_[xx] = yy;
        dist_[xx] = dist_[x] ^ dist_[y] ^ value;
        return true;
    }

    int getDist(int x){ updateDist(x); return dist_[x]; }


private:
    vector<int> fa_;
    vector<int> dist_;
    int sz_;

    inline void updateDist(int x){
        if (fa_[x] == x) { return; }
        int parent = fa_[x];
        fa_[x] = findSet(parent);
        dist_[x] ^= dist_[parent];
        return;
    }
};

void solve(){
    int n, m;
    cin >> n >> m;

    set<int> sett;
    vector<pair<int, int>> a(m);
    vector<int> op(m);
    vector<int> b(2 * m, 0);
    for (int i = 0; i < m; ++i){
        cin >> a[i].first >> a[i].second;
        string s;
        cin >> s;
        op[i] = static_cast<int>(s[0] == 'o');
        b[i * 2] = a[i].first;
        b[i * 2 + 1] = a[i].second;
    }

    sort(b.begin(), b.end());
    b.erase(unique(b.begin(), b.end()), b.end());

    DisjointSet dsu(b.size() + 1);
    int index = 0;
    for (int i = 0; i < m; ++i){
        const auto& [x, y] = a[i];
        int l = lower_bound(b.begin(), b.end(), x - 1) - b.begin();
        int r = lower_bound(b.begin(), b.end(), y) - b.begin();
        if (dsu.unionSet(l, r, op[i])){
            continue;
        }
        else if ((dsu.getDist(l) ^ dsu.getDist(r)) != op[i]){
            cout << i << '\n';
            return;
        }
    }

    cout << m << endl;
}