2019-2020 ACM-ICPC Latin American Regional Programming Contest

开学前最后一场训练了，努力搞了个8题，还算可以。就是这场题没多少我能做的，只切了两道题。

D：

solver：czq

题意就是二维平面上有n颗恒星，每颗恒星有一个固定亮度。问：是否存在一条直线，从平面的一端扫到另一端，满足：对于任意两颗恒星S和T，如果S亮度大于T，那么S一定在T之前或者跟T一起同时被直线扫到。只需输出Y或N。

构造一下Y和N的例子你就会发现，对于所有亮度不同的恒星，从亮度高的恒星指向亮度低的恒星，这样我们就得到若干条向量。若答案为Y，那么必然存在一对向量，平行且方向相反，而且如果我们把所有的向量极角排序，这对平行的向量位置相邻。

 1 /* basic header */
 2 #include <bits/stdc++.h>
 3 /* define */
 4 #define ll long long
 5 #define pb emplace_back
 6 #define mp make_pair
 7 #define eps 1e-8
 8 #define lson (curpos<<1)
 9 #define rson (curpos<<1|1)
10 /* namespace */
11 using namespace std;
12 /* header end */
13 
14 const double pi = acos(-1);
15 const int maxn = 2020;
16 
17 struct Star {
18     int x, y, b;
19 } star[maxn];
20 vector<double>k;
21 int n;
22 
23 double getRad(int x, int y) {
24     if (!y) return x > 0 ? 0 : pi;
25     if (!x) return y > 0 ? pi / 2 : -pi / 2;
26     double ans = atan(double(y) / x);
27     return x > 0 ? ans : ans + pi;
28 }
29 
30 bool check() {
31     if (k.size() < 2) return true;
32     double last = k.back() - pi * 2;
33     for (auto x : k) {
34         if (pi - x + last <= eps) return true;
35         last = x;
36     }
37     return false;
38 }
39 
40 int main() {
41     scanf("%d", &n);
42     for (int i = 0; i < n; i++) {
43         scanf("%d%d%d", &star[i].x, &star[i].y, &star[i].b);
44         for (int j = 0; j < i; j++) {
45             if (star[i].b > star[j].b) k.pb(getRad(star[i].x - star[j].x, star[i].y - star[j].y));
46             else if (star[i].b < star[j].b) k.pb(getRad(star[j].x - star[i].x, star[j].y - star[i].y));
47         }
48     }
49     sort(k.begin(), k.end());
50     if (check()) puts("Y"); else puts("N");
51     return 0;
52 }

View Code

E：

solver：lzh

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 #define ff first
 4 #define ss second
 5 typedef long long ll;
 6 typedef pair<int, int> pii;
 7 
 8 char s[200010];
 9 int a[200010];
10 int main() {
11     int m;
12     scanf("%s%d", s + 1, &m);
13     int n = strlen(s + 1);
14     for (int i = 1; i <= n; i++)
15         s[i + n] = s[i];
16     for (int i = 1; i <= 2 * n; i++)
17         if (s[i] == 'E')
18             a[++a[0]] = i;
19     if (!a[0]) {
20         printf("0
");
21         return 0;
22     }
23     int last = 1;
24     ll ans = 0;
25     for (int i = 1; i <= n; i++) {
26         if (a[last] < i && last <= a[0])
27             last++;
28         if (last > a[0])
29             break;
30         ans += max(0, i + m - a[last]);
31     }
32     printf("%lld
", ans);
33 }

View Code

F：

solver：lzh、zyh

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 #define ff first
 4 #define ss second
 5 typedef long long ll;
 6 typedef pair<int, int> pii;
 7 
 8 ll a[5010][5010];
 9 const int mod = 1e9 + 7;
10 int main() {
11     int n, m;
12     scanf("%d%d", &n, &m);
13     if (n == 1)
14         printf("1
");
15     else if (n == 2)
16         printf("%d
", m - 1);
17     else {
18         for (int i = 1; i <= 5000; i++)
19             a[1][i] = 1, a[2][i] = i - 1;
20         for (int i = 3; i <= n; i++)
21             for (int j = i; j <= m; j++) {
22                 int tmp = j - 1;
23                 while (tmp > 0) {
24                     a[i][j] = (a[i][j] + a[i - 1][tmp] * 2 % mod - a[i - 2][tmp - 1] + mod) % mod;
25                     tmp -= i;
26                 }
27             }
28         printf("%lld
", a[n][m]);
29     }
30 }

View Code

G：

solver：zyh、czq

 1 #include <iostream>
 2 #include <cstring>
 3 #include <algorithm>
 4 #include <vector>
 5 using namespace std;
 6 #define STRMAXLEN 500000
 7 struct SAM {
 8     struct node {
 9         int parent, next[26], len;
10         int cnt;
11     };
12     node tr[STRMAXLEN];
13     int sz[STRMAXLEN];
14     int rnk[STRMAXLEN];
15     int last = 0;
16     int num = 0;
17     void init() {
18         last = 0;
19         num = 0;
20         tr[0].parent = -1;
21     }
22     void add(char ch) {
23         int p = last;
24         int t = ch - 'A';
25         int now = ++num;
26         tr[now].len = tr[last].len + 1;
27         tr[now].cnt = 1;
28         while (p != -1 && tr[p].next[t] == 0) {
29             tr[p].next[t] = now;
30             p = tr[p].parent;
31         }
32         if (p != -1) {
33             int q = tr[p].next[t];
34             if (tr[p].len + 1 == tr[q].len) {
35                 tr[num].parent = q;
36             } else {
37                 tr[++num].parent = tr[q].parent;
38                 memcpy(tr[num].next, tr[q].next, sizeof(tr[num].next));
39                 tr[num].len = tr[p].len + 1;
40                 while (p != -1 && tr[p].next[t] == q) {
41                     tr[p].next[t] = num;
42                     p = tr[p].parent;
43                 }
44                 tr[now].parent = tr[q].parent = num;
45             }
46         } else tr[now].parent = 0;
47         last = now;
48     }
49     void solve() {
50         //依赖排序，可以按这个顺序dp
51         for (int i = 0; i <= num; ++i) sz[i] = 0;
52         for (int i = 0; i <= num; ++i) sz[tr[i].len]++;
53         for (int i = 1; i <= num; ++i) sz[i] += sz[i - 1];
54         for (int i = 0; i <= num; ++i) rnk[--sz[tr[i].len]] = i;
55         for (int i = num; i >= 1; --i) {
56             tr[tr[rnk[i]].parent].cnt += tr[rnk[i]].cnt;
57         }
58     }
59     int getAns(char ch[], int len) {
60         int ans = 1;
61         int now = 0;
62         for (int i = 0; i < len; ++i) {
63             now = tr[now].next[ch[i] - 'A'];
64             if (now == 0) {
65                 ++ans;
66                 now = tr[now].next[ch[i] - 'A'];
67                 if (now == 0) return -1;
68             }
69             //cout<<now<<' ';
70         }
71         //cout<<endl;
72         return ans;
73     }
74 };
75 SAM sam;
76 char str[500001];
77 int main() {
78     scanf("%s", str);
79     sam.init();
80     int len = strlen(str);
81     for (int i = 0; i < len; ++i) sam.add(str[i]);
82     int n;
83     scanf("%d", &n);
84     while (n--) {
85         scanf("%s", str);
86         len = strlen(str);
87         printf("%d
", sam.getAns(str, len));
88     }
89 }

View Code

I：

solver：czq

 1 /* basic header */
 2 #include <bits/stdc++.h>
 3 /* define */
 4 #define ll long long
 5 #define pb emplace_back
 6 #define mp make_pair
 7 #define eps 1e-8
 8 #define lson (curpos<<1)
 9 #define rson (curpos<<1|1)
10 /* namespace */
11 using namespace std;
12 /* header end */
13 
14 const int maxn = 2020;
15 const ll mod = 1e9 + 7;
16 vector<ll>edge[maxn];
17 ll n, m, dp[maxn], cnt = 0, ans = 0;
18 
19 ll dfs(ll cur) {
20     if (dp[cur] != -1) return dp[cur];
21     if (cur > m) {
22         cnt++;
23         dp[cur] = 1;
24         return dp[cur];
25     }
26     ll sum = 0;
27     for (auto x : edge[cur]) sum = ((dfs(x) % mod) + sum) % mod;
28     dp[cur] = sum;
29     return sum;
30 }
31 
32 int main() {
33     for (int i = 0; i < maxn; i++) dp[i] = -1;
34     scanf("%lld%lld", &n, &m);
35     for (int i = 1; i <= m; i++) {
36         ll x; scanf("%lld", &x);
37         while (x--) {
38             ll y; scanf("%lld", &y);
39             edge[i].pb(y);
40         }
41         sort(edge[i].begin(), edge[i].end());
42     }
43     ll ans = dfs(1);
44     printf("%lld %lld
", ans, cnt);
45     return 0;
46 }

View Code

K：

solver：lzh、zyh、czq

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 #define ff first
 4 #define ss second
 5 typedef long long ll;
 6 typedef pair<int, int> pii;
 7 
 8 char s[10010];
 9 int main() {
10     scanf("%s", s + 1);
11     int n = strlen(s + 1);
12     vector<ll> ans, noteql;
13     for (int i = 1; i < n; i++)
14         if (s[i] != s[i + 1])
15             noteql.push_back(2 * i + 1);
16     printf("%d
", noteql.size());
17     if (noteql.size() == 0) {
18         if (s[1] == 'A')
19             printf("-1
");
20         else
21             printf("1
");
22         return 0;
23     }
24     ans.push_back(1), ans.push_back(-noteql.back()), noteql.pop_back();
25 
26     for (auto i : noteql) {
27         vector<ll> tmp = ans;
28         tmp.push_back(0);
29         for (int j = 1; j < tmp.size(); j++)
30             tmp[j] -= i * ans[j - 1];
31         ans = tmp;
32     }
33     ll f = 1;
34     if (((noteql.size() + 1) % 2 == 0 && s[1] == 'A') || ((noteql.size() + 1) % 2 == 1 && s[1] == 'H'))
35         f = -1;
36     // for (auto i : ans)
37     // printf("%d ", f * i);
38     for (int i = 0; i < (int)ans.size(); i++) {
39         printf("%lld%c", f * ans[i], i == (int)ans.size() - 1 ? '
' : ' ');
40     }
41     return 0;
42 }

View Code

L：

solver：lzh、zyh

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 #define ff first
 4 #define ss second
 5 typedef long long ll;
 6 typedef pair<int, int> pii;
 7 
 8 char s[1010][1010];
 9 int a[1010][1010], b[1010];
10 int d[20][10000];
11 int query(int l, int r) {
12     int t = log2(r - l + 1);
13     return min(d[t][l], d[t][r - (1 << t) + 1]);
14 }
15 void init(int len) {
16     for (int i = 0; i < len; ++i) {
17         d[0][i] = b[i];
18     }
19     int t = 1;
20     for (int i = 1; t <= len; ++i) {
21         for (int j = 0; j + t < len; ++j)
22             d[i][j] = min(d[i - 1][j], d[i - 1][j + t]);
23         t <<= 1;
24     }
25 }
26 int main() {
27     int n, m, ans = 0;
28     scanf("%d%d", &n, &m);
29     for (int i = 1; i <= n; i++) {
30         scanf("%s", s[i] + 1);
31         for (int j = 1; j <= m; j++) {
32             if (s[i][j] == 'G')
33                 a[i][j]++;
34             a[i][j] += a[i][j - 1];
35         }
36     }
37     for (int j = 1; j <= m; j++) {
38         for (int i = 1; i <= n; i++) {
39             int l = j, r = m, mid, tmp = 0;
40             while (l <= r) {
41                 mid = l + r >> 1;
42                 if (a[i][mid] - a[i][j - 1] == 0 || a[i][mid] - a[i][j - 1] == mid - j + 1)
43                     l = mid + 1, tmp = max(tmp, mid);
44                 else
45                     r = mid - 1;
46             }
47             b[i - 1] = tmp - j + 1;
48         }
49         init(n);
50         for (int i = 1; i <= n; i++) {
51             int l = 1, r = n - i + 1, mid, tmp = 1;
52             while (l <= r) {
53                 mid = l + r >> 1;
54                 if (query(i - 1, i + mid - 2) >= mid)
55                     l = mid + 1, tmp = max(tmp, mid);
56                 else
57                     r = mid - 1;
58             }
59             ans = max(ans, tmp);
60         }
61     }
62     printf("%d
", ans * ans);
63 }

View Code

M：

solver：lzh

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 #define ff first
 4 #define ss second
 5 typedef long long ll;
 6 typedef pair<int, int> pii;
 7 
 8 int a[1010];
 9 int main() {
10     int n, x;
11     cin >> n >> x;
12     for (int i = 1; i <= n; i++)
13         cin >> a[i];
14     int ans = 1;
15     for (int i = 1; i <= n; i++) {
16         int tmp = 1;
17         for (int j = i + 1; j <= n; j++)
18             if (a[j] - a[j - 1] <= x)
19                 tmp++;
20             else
21                 break;
22         ans = max(ans, tmp);
23     }
24     cout << ans << endl;
25 }

View Code

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三大自由之二部分
原文地址：https://www.cnblogs.com/JHSeng/p/12347673.html