题目链接

D:
将第一种关系建图，求出图上每个点，满足不相邻但可达且不含第二种关系的点的数量，直接建图，注意减数量的时候不要重复即可

#include<bits/stdc++.h>
using namespace std;
#define ms(x,y) memset(x, y, sizeof(x))
#define lowbit(x) ((x)&(-x))
#define sqr(x) ((x)*(x))
typedef long long LL;
typedef pair<int,int> pii;
typedef pair<LL,LL> pll;
#define endl "\n";


void run_case() {
    int n, m, k;
    cin >> n >> m >> k;
    vector<vector<int>> G(n+1, vector<int>());
    vector<unordered_set<int>>a(n+1);
    vector<int> fa(n+1), siz(n+1), son(n+1), ans(n+1);
    for(int i = 0; i < m; ++i) {
        int x, y;
        cin >> x >> y;
        G[x].push_back(y);
        G[y].push_back(x);
        son[x]++, son[y]++;
        a[x].insert(y), a[y].insert(x);
    }
    function<void(int,int,int&)> dfs = [&](int u, int f, int& cnt) {
        fa[u] = f; cnt++;
        for(auto v: G[u])
            if(!fa[v]) dfs(v, f, cnt);
    };
    for(int i = 1; i <= n; ++i) {
        if(!fa[i]) {
            int cnt = 0;
            dfs(i, i, cnt);
            siz[i] = cnt;
        }
    }
    for(int i = 1; i <= n; ++i)
        ans[i] = siz[fa[i]] - son[i] - 1;
    for(int i = 0; i < k; ++i) {
        int x, y;
        cin >> x >> y;
        if(fa[x] == fa[y] && !a[x].count(y)) ans[x]--, ans[y]--;
    }
    for(int i = 1; i <= n; ++i) cout << ans[i] << " ";
}


int main() {
    ios::sync_with_stdio(false), cin.tie(0);
    cout.flags(ios::fixed);cout.precision(9);
    //int t; cin >> t;
    //while(t--)
    run_case();
    cout.flush();
    return 0;
}

F:
在线查询与修改，数据范围5e5，那么对26个字母分别维护一个树状数组即可，难度不大

#include<bits/stdc++.h>
using namespace std;
#define ms(x,y) memset(x, y, sizeof(x))
#define lowbit(x) ((x)&(-x))
#define sqr(x) ((x)*(x))
typedef long long LL;
typedef pair<int,int> pii;
typedef pair<LL,LL> pll;
#define endl "\n";

const int maxn = 5e5+7;

LL a[maxn][26];

void add(int pos, int ch, int val) {
    for(;pos<maxn;pos+=lowbit(pos))
        a[pos][ch] += val;
}

LL getsum(int pos, int ch) {
    LL ret = 0;
    for(;pos; pos -= lowbit(pos))
        ret += a[pos][ch];
    return ret;
}

void run_case() {
    int n, q; string s;
    cin >> n >> s >> q;
    for(int i = 0; i < n; ++i) {
        add(i+1, s[i]-'a', 1);
    }
    while(q--) {
        static int type;
        cin >> type;
        if(type == 1) {
            static int pos;
            static char ch;
            cin >> pos >> ch;
            add(pos, s[pos-1]-'a', -1);
            s[pos-1] = ch;
            add(pos, ch-'a', 1);
        } else {
            static int l, r;
            cin >> l >> r;
            int ans = 0;
            for(int i = 0; i < 26; ++i) 
                ans += ((getsum(r, i)-getsum(l-1,i))>0?1:0);
            cout << ans << endl;
        }
    }
}


int main() {
    ios::sync_with_stdio(false), cin.tie(0);
    cout.flags(ios::fixed);cout.precision(9);
    //int t; cin >> t;
    //while(t--)
    run_case();
    cout.flush();
    return 0;
}