Codeforces Edu Round 67 (Rated for Div. 2)

题目质量不错的一场。

题目链接：https://codeforces.com/contest/1187

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A：

水题。

/* basic header */
#include <bits/stdc++.h>
/* define */
#define ll long long
#define dou double
#define pb emplace_back
#define mp make_pair
#define sot(a,b) sort(a+1,a+1+b)
#define rep1(i,a,b) for(int i=a;i<=b;++i)
#define rep0(i,a,b) for(int i=a;i<b;++i)
#define eps 1e-8
#define int_inf 0x3f3f3f3f
#define ll_inf 0x7f7f7f7f7f7f7f7f
#define lson curpos<<1
#define rson curpos<<1|1
/* namespace */
using namespace std;
/* header end */

int q;

int main() {
    cin >> q;
    while (q--) {
        int n, s, t; cin >> n >> s >> t;
        int chong = s + t - n; s -= chong, t -= chong;
        cout << max(s, t) + 1 << endl;
    }
    return 0;
}

B：

用数据结构维护s的每个字母出现的位置即可。

tag里的binary search怪怪的，我觉得ds才靠谱(

/* basic header */
#include <bits/stdc++.h>
/* define */
#define ll long long
#define dou double
#define pb emplace_back
#define mp make_pair
#define sot(a,b) sort(a+1,a+1+b)
#define rep1(i,a,b) for(int i=a;i<=b;++i)
#define rep0(i,a,b) for(int i=a;i<b;++i)
#define eps 1e-8
#define int_inf 0x3f3f3f3f
#define ll_inf 0x7f7f7f7f7f7f7f7f
#define lson curpos<<1
#define rson curpos<<1|1
/* namespace */
using namespace std;
/* header end */

vector<int>v[26];
string a, b;
int n, m, p = 0;

int main() {
    cin >> n >> a;
    for (auto i : a) v[i - 'a'].pb(++p);
    cin >> m;
    while (m--) {
        cin >> b;
        int cnt[26] = {0};
        for (auto i : b) cnt[i - 'a']++;
        int ans = 0;
        rep0(i, 0, 26)
        if (cnt[i]) ans = max(ans, v[i][cnt[i] - 1]);
        printf("%d
", ans);
    }
    return 0;
}

C：

考察每个区间，如果要求不下降，则给区间内每个成员打标记。然后看下降区间内有无未被标记成员，若无，则不可能构造。最后构造出来的数组不下降区间内所有成员大小相同即可，满足数组整体不上升。

/* basic header */
#include <bits/stdc++.h>
/* define */
#define ll long long
#define dou double
#define pb emplace_back
#define mp make_pair
#define sot(a,b) sort(a+1,a+1+b)
#define rep1(i,a,b) for(int i=a;i<=b;++i)
#define rep0(i,a,b) for(int i=a;i<b;++i)
#define eps 1e-8
#define int_inf 0x3f3f3f3f
#define ll_inf 0x7f7f7f7f7f7f7f7f
#define lson curpos<<1
#define rson curpos<<1|1
/* namespace */
using namespace std;
/* header end */

const int maxn = 1e3 + 10;
struct Interval {
    int t, l, r;
    Interval() {}
    Interval(int a, int b, int c): t(a), l(b), r(c) {}
} v[maxn];
int n, m, p = 0, ans = 1e8, flag = 1, a[maxn], b[maxn];

int main() {
    scanf("%d%d", &n, &m);
    rep1(i, 1, m) {
        scanf("%d%d%d", &v[i].t, &v[i].l, &v[i].r);
        if (v[i].t == 1)
            rep0(j, v[i].l, v[i].r) a[j] = 1; // sign
    }
    rep1(i, 1, m)
    if (!v[i].t) {
        flag = 0;
        rep0(j, v[i].l, v[i].r)
        if (!a[j]) {
            flag = 1; break;
        }
        if (!flag)
            return puts("NO");
    }
    puts("YES");
    rep1(i, 1, n) printf("%d ", a[i - 1] ? ans : --ans);
    puts("");
    return 0;
}

D：

用线段树维护数组a的区间最小值。另开一个vector<int>[]倒序记录数组中每个数字的出现位置。

O(n)遍历数组b，对于每一个b[i]，检查从1到b[i]最先一次出现在a数组的位置(这个位置每使用一次会往后更新，看代码注释)这段区间内的最小值是否等于b[i]。若是，则有解，反之无解。

/* basic header */
#include <bits/stdc++.h>
/* define */
#define ll long long
#define dou double
#define pb emplace_back
#define mp make_pair
#define sot(a,b) sort(a+1,a+1+b)
#define rep1(i,a,b) for(int i=a;i<=b;++i)
#define rep0(i,a,b) for(int i=a;i<b;++i)
#define eps 1e-8
#define int_inf 0x3f3f3f3f
#define ll_inf 0x7f7f7f7f7f7f7f7f
#define lson (curpos<<1)
#define rson (curpos<<1|1)
#define mid (curl+curr>>1)
/* namespace */
using namespace std;
/* header end */

const int maxn = 3e5 + 10;
int t, n, segt[maxn << 2], a[maxn], b[maxn];
vector<int>pos[maxn];

void build(int curpos, int curl, int curr) {
    segt[curpos] = a[curl];
    if (curl == curr) return;
    build(lson, curl, mid); build(rson, mid + 1, curr);
    segt[curpos] = min(segt[lson], segt[rson]);
}

int query(int curpos, int curl, int curr, int ql, int qr) {
    if (ql <= curl && curr <= qr) return segt[curpos];
    if (qr <= mid) return query(lson, curl, mid, ql, qr);
    else if (ql > mid) return query(rson, mid + 1, curr, ql, qr);
    else return min(query(lson, curl, mid, ql, mid), query(rson, mid + 1, curr, mid + 1, qr));
}

void update(int curpos, int curl, int curr, int pos, int val) {
    if (curl == curr) {
        segt[curpos] = val;
        return;
    }
    if (pos <= mid) update(lson, curl, mid, pos, val);
    else update(rson, mid + 1, curr, pos, val);
    segt[curpos] = min(segt[lson], segt[rson]);
}

int main() {
    scanf("%d", &t);
    while (t--) {
        scanf("%d", &n);
        rep1(i, 1, n) pos[i].clear();
        rep1(i, 1, n) scanf("%d", &a[i]);
        for (int i = n; i > 0; i--) pos[a[i]].pb(i);
        rep1(i, 1, n) scanf("%d", &b[i]);
        int flag = 1;
        build(1, 1, n);
        rep1(i, 1, n) {
            int k = b[i];
            if (pos[k].empty()) { // no element can be choisen to swap
                flag = 0; break;
            }
            int x = *(--pos[k].end()); // get the last element of vector pos[k]
            if (query(1, 1, n, 1, x) != k) { // the mininum of interval is not current b[i], no solution obviously
                flag = 0; break;
            }
            update(1, 1, n, x, int_inf); // current b[i] is ok, make it into inf to eliminate the affect
        }
        if (flag) puts("YES"); else puts("NO");
    }
    return 0;
}

E：

答案就是每个点的总度数+点数。跑一遍dfs即可。

/* basic header */
#include <bits/stdc++.h>
/* define */
#define ll long long
#define dou double
#define pb emplace_back
#define mp make_pair
#define sot(a,b) sort(a+1,a+1+b)
#define rep1(i,a,b) for(int i=a;i<=b;++i)
#define rep0(i,a,b) for(int i=a;i<b;++i)
#define eps 1e-8
#define int_inf 0x3f3f3f3f
#define ll_inf 0x7f7f7f7f7f7f7f7f
#define lson (curpos<<1)
#define rson (curpos<<1|1)
#define mid (curl+curr>>1)
/* namespace */
using namespace std;
/* header end */

const int maxn = 5e5 + 10;
int n, top = 0, maxx, pos, cur;
ll ans = 0;
int head[maxn], nxt[maxn], edge[maxn], depth[maxn], size[maxn];

void addEdge(int x, int y) {
    edge[++top] = y;
    nxt[top] = head[x];
    head[x] = top;
}

void dfs(int x, int fa) {
    size[x] = 1;
    for (int i = head[x]; i; i = nxt[i]) {
        int v = edge[i];
        if (v == fa) continue;
        depth[v] = depth[x] + 1;
        dfs(v, x);
        size[x] += size[v];
    }
}

void calc(int x, int fa, ll num) {
    ans = max(ans, num);
    for (int i = head[x]; i; i = nxt[i]) {
        int v = edge[i];
        if (v == fa) continue;
        calc(v, x, num + n - 2 * size[v]);
    }
}

int main() {
    scanf("%d", &n);
    rep0(i, 1, n) {
        int x, y; scanf("%d%d", &x, &y);
        addEdge(x, y); addEdge(y, x);
    }
    depth[1] = 1;
    dfs(1, 0);
    ll num = 0;
    rep1(i, 1, n) num += depth[i];
    calc(1, 0, num);
    printf("%lld
", ans);
    return 0;
}

F && G：

不会，摸了(