Avito Cool Challenge 2018 Solution

A. Definite Game

签.

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3
 4 int main()
 5 {
 6     int a;
 7     while (scanf("%d", &a) != EOF)
 8     {
 9         [](int x)
10         {
11             for (int i = x - 1; i >= 1; --i) if (x % i)
12             {
13                 printf("%d\n", x - i);
14                 return;
15             }
16             printf("%d\n", x);
17         }(a);
18     }
19     return 0;
20 }

B. Farewell Party

签。

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3
 4 #define N 100010
 5 int n, a[N], b[N], cnt[N];
 6 vector <int> v[N];
 7 int ans[N];
 8
 9 bool ok()
10 {
11     for (int i = 1; i <= n; ++i) if (!v[i].empty() && v[i].size() != i)
12         return false;
13     return true;
14 }
15 void solve()
16 {
17     int cnt = 0;
18     for (int i = 1; i <= n; ++i) v[i].clear();
19     for (int i = 1; i <= n; ++i)
20     {
21         int id = n - a[i];
22         if (v[id].size() == id)
23         {
24             ++cnt;
25             for (auto it : v[id]) ans[it] = cnt;
26             v[id].clear();
27         }
28         v[id].push_back(i);
29     }
30     if (!ok())
31     {
32         puts("Impossible");
33         return;
34     }
35     else
36     {
37         for (int i = 1; i <= n; ++i) if (!v[i].empty())
38         {
39             ++cnt;
40             for (auto it : v[i])
41                 ans[it] = cnt;
42         }
43         puts("Possible");
44         for (int i = 1; i <= n; ++i) printf("%d%c", ans[i], " \n"[i == n]);
45     }
46 }
47
48 int main()
49 {
50     while (scanf("%d", &n) != EOF)
51     {
52         for (int i = 1; i <= n; ++i) scanf("%d", a + i);
53         solve();
54     }
55     return 0;
56 }

D. Maximum Distance

Upsolved.

题意：

有一张图

定义一条路径的长度的路径上边权最大值

定义两点距离为两点之间最短路径

有一些特殊点，要对所有特殊点求离它最远的特殊点

思路：

我们考虑一条边所连接的两个连通块里，如果这两个连通块里都有特殊点

那么这条边的权值就可以用于更新特殊点的答案

其实就是一个求最小生成树的过程，先按边权排序

要注意的是不是求整个图的最小生成树，而是所有特殊点的最小生成树

其实是最小瓶颈生成树.

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3
 4 #define N 100010
 5 int n, m, k;
 6 struct node
 7 {
 8     int u, v, w;
 9     void scan() { scanf("%d%d%d", &u, &v, &w); }
10     bool operator < (const node &other) const { return w < other.w; }
11 }edge[N];
12
13 int pre[N], cnt[N];
14 int find(int x) { return pre[x] == 0 ? x : pre[x] = find(pre[x]); }
15 int Kruskal()
16 {
17     sort(edge + 1, edge + 1 + m);
18     for (int i = 1; i <= m; ++i)
19     {
20         int u = edge[i].u, v = edge[i].v, w = edge[i].w;
21         int fu = find(u), fv = find(v);
22         if (fu == fv) continue;
23         cnt[fv] += cnt[fu];
24         pre[fu] = fv;
25         if (cnt[fv] == k) return w;
26     }
27 }
28
29 int main()
30 {
31     while (scanf("%d%d%d", &n, &m, &k) != EOF)
32     {
33         memset(pre, 0, sizeof pre);
34         memset(cnt, 0, sizeof cnt);
35         for (int i = 1, x; i <= k; ++i) scanf("%d", &x), cnt[x] = 1;
36         for (int i = 1; i <= m; ++i) edge[i].scan();
37         int res = Kruskal();
38         for (int i = 1; i <= k; ++i) printf("%d%c", res, " \n"[i == k]);
39     }
40     return 0;
41 }

原文地址：https://www.cnblogs.com/Dup4/p/10332497.html