A题

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

typedef long long ll;

typedef pair<int, int> pii;
#define fst first
#define snd second

const int MAXN = 100010;

int main(void) {
    int n;
    cin >> n;
    vector<int> st;
    for (int i = 0; i < n; ++ i) {
        int x;
        cin >> x;
        if (!st.empty() && (st.back() + x) % 2 == 1) st.pop_back();
        else st.push_back(x);
    }
    cout << st.size() << endl;

    return 0;
}

===================================

B题

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

typedef long long ll;
typedef pair<int, int> pii;
#define fst first
#define snd second

const int MAXN = 100010;
bool can[30][30];
int dp[100010][30];

int main(void) {

    int n;
    cin >> n;
    string s;
    cin >> s;

    for (int i = 0; i < 26; ++ i) for (int j = 0; j < 26; ++ j) can[i][j] = true;
    int m;
    cin >> m;
    for (int i = 0; i < m; ++ i) {
        string t;
        cin >> t;
        can[t[0]-'a'][t[1]-'a'] = false;
        can[t[1]-'a'][t[0]-'a'] = false;
    }

    memset(dp, 0x3f, sizeof(dp));
    vector<int> lst(26, 0);
    lst[s[0] - 'a'] = 1;
    dp[1][s[0] - 'a'] = 0;
    for (int i = 2; i <= n; ++ i) {
        for (int j = 0; j < 26; ++ j) {
            dp[i][j] = dp[i - 1][j] + 1;
        }

        int ch = s[i - 1] - 'a';
        for (int j = 0; j < 26; ++ j) if (can[ch][j] && lst[j]) {
            dp[i][ch] = min(dp[i][ch], dp[lst[j]][j] + i - lst[j] - 1);
        }
        dp[i][ch] = min(dp[i][ch], i - 1);

        lst[ch] = i;
    }

    int ans = n;
    for (int i = 0; i < 26; ++ i) ans = min(ans, dp[n][i]);
    cout << ans << endl;

    return 0;
}

===================== C题

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

typedef long long ll;
typedef pair<ll, ll> pii;
#define fst first
#define snd second

const int MAXN = 100010;
const ll inf = 1000000000000LL;

vector<pii> ow[10010];
//priority_queue<pii, vector<pii>, greater<pii>> pq[110];
ll e[10010];
int lst[10010];

int main(void) {

    map<int, int> sid;
    int n, m, K;
    cin >> n >> m >> K;
    //for (int i = 1; i <= m; ++ i) e[i].fst = inf;
    vector<ll> s(n), t(n), p(n);
    for (int i = 0; i < n; ++ i) {
        lst[i] = 0;
        cin >> s[i] >> t[i] >> p[i];
        sid[s[i]] = i;
        e[i] = t[i] + K;
        for (int j = 0; j < p[i]; ++ j) {
            ll o, w;
            cin >> o >> w;
            ow[i].emplace_back(pii(o, w));
            /*
            if (j == 0 && (t[i] + K < e[o].fst || (t[i] + K == e[o].fst && e[o].snd > s[i]))) {
                e[o].fst = t[i] + K;
                e[o].snd = s[i];
            }
            */
        }
    }

    priority_queue<pii, vector<pii>, greater<pii>> pq;
    for (int i = 0; i < n; ++ i) pq.push(pii(e[i], s[i]));

    vector<ll> curt(m + 1, 0);
    vector<ll> ans(n);
    while ( !pq.empty() ) {
        ll ee = inf, eid = 0;
        /*
        for (int i = 0; i < n; ++ i) if (lst[i] < p[i]) {
            if (e[i] < ee || (e[i] == ee && s[eid] > s[i])) {
                eid = i;
                ee = e[i];
            }
        }
        */
        /*
        for (int i = 1; i <= m; ++ i) {
            if (!pq[i].empty() && pq[i].top().fst < ee) {
                ee = pq[i].top().fst;
                eid = i;
            }
        }
        */
        ee = pq.top().fst, eid = sid[pq.top().snd];
        pq.pop();
        //if (ee == inf) break;

        int id = ow[eid][lst[eid]].fst;
        //pii cur = pq[eid].top(); pq[eid].pop();
        //ll time = cur.fst;
        //ll id = sid[cur.snd];
        //cerr << eid << ' ' << id << ' ' << e[eid] << endl;
        if (lst[eid] >= (ll)ow[eid].size() - 1) {
            ans[eid] = max(e[eid], curt[id]) + ow[eid][lst[eid]].snd;
            e[eid] = ans[eid];
            curt[id] = ans[eid];
        } else {
            //ll nid = ow[eid][lst[eid] + 1].fst;
            ll ntime = max(e[eid], curt[id]) + ow[eid][lst[eid]].snd + K;
            curt[id] = ntime - K;
            e[eid] = ntime;
            pq.push(pii(ntime, s[eid]));
        }
        ++ lst[eid];
    }

    for (int i = 0; i < n; ++ i) cout << ans[i] << endl;

    return 0;
}