2018-2019 ICPC Northwestern European Regional Programming Contest (NWERC 2018)

J题队友犯了初始化的错，白给6发，本来能1A的

B：

solver：lzh、czq

就是个拓扑排序

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 #define ff first
 4 #define ss second
 5 #define mp make_pair
 6 typedef long long ll;
 7 typedef pair<int, int> pii;
 8 
 9 vector<int> g[400010];
10 vector<int> v;
11 int a[400010], deg[400010];
12 int main() {
13     v.push_back(0);
14     int n;
15     scanf("%d", &n);
16     for (int i = 0; i < n; i++) {
17         int j;
18         scanf("%d%d", &a[i], &j);
19         while (j--) {
20             int x;
21             scanf("%d", &x);
22             g[i].push_back(x - 1);
23             deg[x - 1]++;
24         }
25     }
26     int ans = 0;
27     priority_queue<pii> pq;
28     for (int i = 0; i < n; i++)
29         if (!deg[i]) pq.push(mp(-a[i], i));
30     while (!pq.empty()) {
31         pii x = pq.top();
32         pq.pop();
33         v.push_back(-x.ff);
34         for (auto i : g[x.ss]) {
35             deg[i]--;
36             if (!deg[i])
37                 pq.push(mp(-a[i], i));
38         }
39     }
40     reverse(v.begin(), v.end());
41     for (int i = 0; i < n; i++) ans = max(ans, v[i] + i);
42     printf("%d\n", ans);
43 }
View Code

C：

solver：czq

直接把圆周切个1000份

 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 = 1010;
15 const double pi = acos(-1);
16 vector<pair<double, double>>ans;
17 vector<int>g[maxn];
18 int n, __size[maxn];
19 
20 void dfs(int u, int father, double angle) {
21     __size[u]++;
22     for (auto v : g[u]) {
23         if (v != father) {
24             ans[v].first = ans[u].first + cos(angle);
25             ans[v].second = ans[u].second + sin(angle);
26             dfs(v, u, angle);
27             __size[u] += __size[v];
28             angle += pi / maxn * __size[v];
29         }
30     }
31 }
32 
33 int main() {
34     scanf("%d", &n);
35     ans.resize(n + 1);
36     for (int i = 1; i < n; i++) {
37         int u, v; scanf("%d%d", &u, &v);
38         g[u].pb(v), g[v].pb(u);
39     }
40     dfs(1, -1, 0);
41     for (int i = 1; i <= n; i++) printf("%.10f %.10f\n", ans[i].first, ans[i].second);
42     return 0;
43 }
View Code

G：

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[30];
 9 int notuse[30];
10 const int dd[4][2] = { -1, 0, 1, 0, 0, 1, 0, -1 };
11 const int C[4][2] = { 0, 1, 1, 0, 2, 3, 3, 2 };
12 const char fa[] = { "LRUD" };
13 int main() {
14     scanf("%s", s + 1);
15     int n = strlen(s + 1);
16     if (n == 1) {
17         if (s[1] == 'L')
18             printf("1 0\n0\n");
19         else if (s[1] == 'R')
20             printf("-1 0\n0\n");
21         else if (s[1] == 'U')
22             printf("0 -1\n0\n");
23         else
24             printf("0 1\n0\n");
25         return 0;
26     }
27     for (int j = 0; j <= 4; j += 2) {
28         vector<int> tmp;
29         for (int i = 1; i <= n; i++)
30             if (s[i] == fa[j] || s[i] == fa[j + 1])
31                 tmp.push_back(i);
32         for (int i = tmp.size() - 3; i >= 0; i--)
33             if (tmp[i] + 1 == tmp[i + 1] && tmp[i + 1] + 1 == tmp[i + 2])
34                 notuse[tmp[i]]++;
35     }
36     for (int j = 0; j <= 4; j += 2) {
37         int tmp = 0;
38         for (int i = n; i >= 1; i--)
39             if (s[i] == fa[j] || s[i] == fa[j + 1])
40                 tmp++;
41             else
42                 break;
43         if (tmp >= 3) {
44             printf("impossible\n");
45             return 0;
46         }
47     }
48 
49     vector<pii> ans;
50     pii cur = { 0, 0 };
51     int step = n + 1;
52     for (int i = n; i >= 1; i--)
53         if (!notuse[i]) {
54             for (int j = 0; j < 4; j++)
55                 if (s[i] == fa[j]) {
56                     if (i != n)
57                         ans.push_back({ cur.ff + dd[j][0], cur.ss + dd[j][1] });
58 
59                     pii tmp = { (1 << step) *(-dd[j][0]), (1 << step) *(-dd[j][1]) };
60                     if (tmp.ff)
61                         cur.ff += tmp.ff;
62                     else
63                         cur.ss += tmp.ss;
64                 }
65             step--;
66         }
67     printf("%d %d\n%d\n", cur.ff, cur.ss, ans.size());
68     for (auto i : ans)
69         printf("%d %d\n", i.ff, i.ss);
70 }
View Code

H：

solver：zyh、lzh

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 bool z[1000001];
 4 int main() {
 5     int n, c, b;
 6     scanf("%d%d%d", &n, &c, &b);
 7     for (int i = 0; i < b; ++i) {
 8         int t;
 9         scanf("%d", &t);
10         z[t - 1] = true;
11     }
12     int now = 0;
13     if (c & 1) {
14         now = 1;
15         c++;
16     }
17     for (int i = 0; i < n; ++i) {
18         if (z[i] || !c) printf("0");
19         else {
20             printf("%d", now);
21             c--;
22             now = 1 - now;
23         }
24     }
25 }
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 = 2e5 + 10;
15 int n, a[maxn];
16 double ans = INT_MAX;
17 
18 int main() {
19     scanf("%d", &n);
20     for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
21     sort(a + 1, a + 1 + n);
22     for (int i = 1; i <= n; i++) {
23         if (a[i] > i) return puts("-1"), 0;
24         ans = min(ans, double(a[i]) / (i));
25     }
26     printf("%.10f\n", ans);
27     return 0;
28 }
View Code

J：

solver：zyh、czq

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 long long p[200001];
 4 int main() {
 5     int n;
 6     scanf("%d", &n);
 7     for (int i = 1; i <= n; ++i) scanf("%lld", &p[i]);
 8     sort(p + 1, p + n + 1);
 9     p[0] = -1;
10     if (p[n] == p[n - 1]) {
11         printf("%lld\n", p[n] - p[n - 2]); return 0;
12     }
13     long long sum = p[n] - p[n - 1];
14     long long Last = p[n - 1];
15     long long maxT = __LONG_LONG_MAX__;
16     long long T = 1;
17     long long ans = 0;
18     int tot = 0;
19     for (int i = n - 1; i >= 1; --i) {
20         ++tot;
21         if (p[i] != p[i - 1]) {
22             if (maxT >= Last - p[i]) {
23                 ans += T * (Last - p[i]);
24                 sum -= (T - 1) * (Last - p[i]);
25                 T = int(log2(tot)) + 1;
26                 if (T > 1) maxT = sum / (T - 1);
27                 else maxT = sum + 1;
28                 Last = p[i];
29             } else break;
30         }
31     }
32     if (T > 1) ans += T * maxT + sum % (T - 1);
33     else ans += T * maxT;
34     printf("%lld", ans);
35     return 0;
36 }
View Code

K：

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 int n, m;
15 string a, b, tmp, ans = "";
16 
17 int main() {
18     ios::sync_with_stdio(false);
19     cin.tie(0);
20     cin >> n >> m >> a >> b;
21     tmp = a;
22     reverse(b.begin(), b.end());
23     reverse(tmp.begin(), tmp.end());
24     int p = 0;
25     for (int i = 0; i < m - n; i++) {
26         int x = b[i] - tmp[p]; if (x < 0) x += 26;
27         ans += (char)(x + 'a');
28         tmp += ans[p++];
29     }
30     tmp += a;
31     reverse(ans.begin(), ans.end());
32     cout << ans << a << endl;
33     return 0;
34 }
View Code

 

posted @ 2020-02-20 18:24  JHSeng  阅读(989)  评论(0)  编辑  收藏  举报