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

A：

水题。

 1 /* basic header */
 2 #include <bits/stdc++.h>
 3 /* define */
 4 #define ll long long
 5 #define dou double
 6 #define pb emplace_back
 7 #define mp make_pair
 8 #define sot(a,b) sort(a+1,a+1+b)
 9 #define rep1(i,a,b) for(int i=a;i<=b;++i)
10 #define rep0(i,a,b) for(int i=a;i<b;++i)
11 #define eps 1e-8
12 #define int_inf 0x3f3f3f3f
13 #define ll_inf 0x7f7f7f7f7f7f7f7f
14 #define lson curpos<<1
15 #define rson curpos<<1|1
16 /* namespace */
17 using namespace std;
18 /* header end */
19 
20 int q;
21 
22 int main() {
23     cin >> q;
24     while (q--) {
25         int n, s, t; cin >> n >> s >> t;
26         int chong = s + t - n; s -= chong, t -= chong;
27         cout << max(s, t) + 1 << endl;
28     }
29     return 0;
30 }

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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(

 1 /* basic header */
 2 #include <bits/stdc++.h>
 3 /* define */
 4 #define ll long long
 5 #define dou double
 6 #define pb emplace_back
 7 #define mp make_pair
 8 #define sot(a,b) sort(a+1,a+1+b)
 9 #define rep1(i,a,b) for(int i=a;i<=b;++i)
10 #define rep0(i,a,b) for(int i=a;i<b;++i)
11 #define eps 1e-8
12 #define int_inf 0x3f3f3f3f
13 #define ll_inf 0x7f7f7f7f7f7f7f7f
14 #define lson curpos<<1
15 #define rson curpos<<1|1
16 /* namespace */
17 using namespace std;
18 /* header end */
19 
20 vector<int>v[26];
21 string a, b;
22 int n, m, p = 0;
23 
24 int main() {
25     cin >> n >> a;
26     for (auto i : a) v[i - 'a'].pb(++p);
27     cin >> m;
28     while (m--) {
29         cin >> b;
30         int cnt[26] = {0};
31         for (auto i : b) cnt[i - 'a']++;
32         int ans = 0;
33         rep0(i, 0, 26)
34         if (cnt[i]) ans = max(ans, v[i][cnt[i] - 1]);
35         printf("%d
", ans);
36     }
37     return 0;
38 }

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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。

 1 /* basic header */
 2 #include <bits/stdc++.h>
 3 /* define */
 4 #define ll long long
 5 #define dou double
 6 #define pb emplace_back
 7 #define mp make_pair
 8 #define sot(a,b) sort(a+1,a+1+b)
 9 #define rep1(i,a,b) for(int i=a;i<=b;++i)
10 #define rep0(i,a,b) for(int i=a;i<b;++i)
11 #define eps 1e-8
12 #define int_inf 0x3f3f3f3f
13 #define ll_inf 0x7f7f7f7f7f7f7f7f
14 #define lson curpos<<1
15 #define rson curpos<<1|1
16 /* namespace */
17 using namespace std;
18 /* header end */
19 
20 const int maxn = 1e3 + 10;
21 struct Interval {
22     int t, l, r;
23     Interval() {}
24     Interval(int a, int b, int c): t(a), l(b), r(c) {}
25 } v[maxn];
26 int n, m, p = 0, ans = 1e8, flag = 1, a[maxn], b[maxn];
27 
28 int main() {
29     scanf("%d%d", &n, &m);
30     rep1(i, 1, m) {
31         scanf("%d%d%d", &v[i].t, &v[i].l, &v[i].r);
32         if (v[i].t == 1)
33             rep0(j, v[i].l, v[i].r) a[j] = 1; // sign
34     }
35     rep1(i, 1, m)
36     if (!v[i].t) {
37         flag = 0;
38         rep0(j, v[i].l, v[i].r)
39         if (!a[j]) {
40             flag = 1; break;
41         }
42         if (!flag)
43             return puts("NO");
44     }
45     puts("YES");
46     rep1(i, 1, n) printf("%d ", a[i - 1] ? ans : --ans);
47     puts("");
48     return 0;
49 }

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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。

 1 /* basic header */
 2 #include <bits/stdc++.h>
 3 /* define */
 4 #define ll long long
 5 #define dou double
 6 #define pb emplace_back
 7 #define mp make_pair
 8 #define sot(a,b) sort(a+1,a+1+b)
 9 #define rep1(i,a,b) for(int i=a;i<=b;++i)
10 #define rep0(i,a,b) for(int i=a;i<b;++i)
11 #define eps 1e-8
12 #define int_inf 0x3f3f3f3f
13 #define ll_inf 0x7f7f7f7f7f7f7f7f
14 #define lson (curpos<<1)
15 #define rson (curpos<<1|1)
16 #define mid (curl+curr>>1)
17 /* namespace */
18 using namespace std;
19 /* header end */
20 
21 const int maxn = 3e5 + 10;
22 int t, n, segt[maxn << 2], a[maxn], b[maxn];
23 vector<int>pos[maxn];
24 
25 void build(int curpos, int curl, int curr) {
26     segt[curpos] = a[curl];
27     if (curl == curr) return;
28     build(lson, curl, mid); build(rson, mid + 1, curr);
29     segt[curpos] = min(segt[lson], segt[rson]);
30 }
31 
32 int query(int curpos, int curl, int curr, int ql, int qr) {
33     if (ql <= curl && curr <= qr) return segt[curpos];
34     if (qr <= mid) return query(lson, curl, mid, ql, qr);
35     else if (ql > mid) return query(rson, mid + 1, curr, ql, qr);
36     else return min(query(lson, curl, mid, ql, mid), query(rson, mid + 1, curr, mid + 1, qr));
37 }
38 
39 void update(int curpos, int curl, int curr, int pos, int val) {
40     if (curl == curr) {
41         segt[curpos] = val;
42         return;
43     }
44     if (pos <= mid) update(lson, curl, mid, pos, val);
45     else update(rson, mid + 1, curr, pos, val);
46     segt[curpos] = min(segt[lson], segt[rson]);
47 }
48 
49 int main() {
50     scanf("%d", &t);
51     while (t--) {
52         scanf("%d", &n);
53         rep1(i, 1, n) pos[i].clear();
54         rep1(i, 1, n) scanf("%d", &a[i]);
55         for (int i = n; i > 0; i--) pos[a[i]].pb(i);
56         rep1(i, 1, n) scanf("%d", &b[i]);
57         int flag = 1;
58         build(1, 1, n);
59         rep1(i, 1, n) {
60             int k = b[i];
61             if (pos[k].empty()) { // no element can be choisen to swap
62                 flag = 0; break;
63             }
64             int x = *(--pos[k].end()); // get the last element of vector pos[k]
65             if (query(1, 1, n, 1, x) != k) { // the mininum of interval is not current b[i], no solution obviously
66                 flag = 0; break;
67             }
68             update(1, 1, n, x, int_inf); // current b[i] is ok, make it into inf to eliminate the affect
69         }
70         if (flag) puts("YES"); else puts("NO");
71     }
72     return 0;
73 }

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E：

答案就是每个点的总度数+点数。跑一遍dfs即可。

 1 /* basic header */
 2 #include <bits/stdc++.h>
 3 /* define */
 4 #define ll long long
 5 #define dou double
 6 #define pb emplace_back
 7 #define mp make_pair
 8 #define sot(a,b) sort(a+1,a+1+b)
 9 #define rep1(i,a,b) for(int i=a;i<=b;++i)
10 #define rep0(i,a,b) for(int i=a;i<b;++i)
11 #define eps 1e-8
12 #define int_inf 0x3f3f3f3f
13 #define ll_inf 0x7f7f7f7f7f7f7f7f
14 #define lson (curpos<<1)
15 #define rson (curpos<<1|1)
16 #define mid (curl+curr>>1)
17 /* namespace */
18 using namespace std;
19 /* header end */
20 
21 const int maxn = 5e5 + 10;
22 int n, top = 0, maxx, pos, cur;
23 ll ans = 0;
24 int head[maxn], nxt[maxn], edge[maxn], depth[maxn], size[maxn];
25 
26 void addEdge(int x, int y) {
27     edge[++top] = y;
28     nxt[top] = head[x];
29     head[x] = top;
30 }
31 
32 void dfs(int x, int fa) {
33     size[x] = 1;
34     for (int i = head[x]; i; i = nxt[i]) {
35         int v = edge[i];
36         if (v == fa) continue;
37         depth[v] = depth[x] + 1;
38         dfs(v, x);
39         size[x] += size[v];
40     }
41 }
42 
43 void calc(int x, int fa, ll num) {
44     ans = max(ans, num);
45     for (int i = head[x]; i; i = nxt[i]) {
46         int v = edge[i];
47         if (v == fa) continue;
48         calc(v, x, num + n - 2 * size[v]);
49     }
50 }
51 
52 int main() {
53     scanf("%d", &n);
54     rep0(i, 1, n) {
55         int x, y; scanf("%d%d", &x, &y);
56         addEdge(x, y); addEdge(y, x);
57     }
58     depth[1] = 1;
59     dfs(1, 0);
60     ll num = 0;
61     rep1(i, 1, n) num += depth[i];
62     calc(1, 0, num);
63     printf("%lld
", ans);
64     return 0;
65 }

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F && G：

不会，摸了(