2015ACM/ICPC亞洲區瀋陽站 Solution

A - Pattern String

留坑。

 

B - Bazinga

題意：找一個最大的i，使得前i - 1個字符串中至少不是它的子串

思路：暴力找，若是有一個串已經符合條件，就不用往上更新

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 
 4 #define N 510
 5 
 6 int t, n;
 7 int vis[N];
 8 string s[N];
 9 
10 int main()
11 {
12     ios::sync_with_stdio(false);
13     int t;
14     cin >> t;
15     for(int cas = 1; cas <= t; ++cas)
16     {
17         cin >> n;
18         memset(vis, 0, sizeof vis);
19         for(int i = 1; i <= n; ++i) cin >> s[i];
20         int ans = -1;
21         for(int i = 1; i <= n; ++i)
22         {
23             for(int j  = i + 1; j <= n; ++j)
24             {
25                 if(!vis[j])
26                 {
27                     int tmp = s[j].find(s[i]);
28                     if(tmp == -1)
29                     {
30                         vis[j] = 1;
31                         ans = max(ans, j);
32                     }
33                     else break;
34                 }
35             }
36         }
37         cout << "Case #" << cas << ": " << ans << endl;
38     }
39     return 0;
40 }

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C - Minimum Cut-Cut

留坑。

 

D - Pagodas

題意：剛開始集合裏有a, b 兩個數，每次能從集合中選出 x + y  或者 x - y 放進集合中，而且放過的不能放，不能放的輸遊戲

思路：考慮$gcd(x, y), 每次產生的新數確定是gcd(x, y)的倍數$ 因此能放的數總數只有 gcd(x, y) 的倍數的個數

 1 #include<bits/stdc++.h>
 2 
 3 using namespace std;
 4 
 5 int gcd(int a, int b)
 6 {
 7     return b == 0 ? a : gcd(b, a % b);
 8 }
 9 
10 int n, a, b;
11 
12 int main()
13 {
14     int t;
15     scanf("%d", &t);
16     for(int cas = 1; cas <= t; ++cas)
17     {
18         scanf("%d %d %d", &n, &a, &b);
19         int g = gcd(a, b);
20         int tmp = n / g;
21         printf("Case #%d: ", cas);
22         puts(tmp & 1 ? "Yuwgna" : "Iaka");
23     }
24     return 0;
25 }

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E - Efficient Tree

留坑。

 

F - Frogs

題意：有n只青蛙，長度爲m的環，每隻青蛙固定步數往前跳，跳到一個點標記一下，求最後有多少點被標記

思路：先將m分解因子，對於每個arr[i]都和m求gcd，而後將gcd的倍數標記一次，表明須要統計一次它的倍數。最後遍歷一邊每一個因子，求出它被多用幾回或者少用幾回的和便可。

 1 #include<bits/stdc++.h>
 2 
 3 using namespace std;
 4 
 5 #define ll long long
 6 #define N 10010
 7 
 8 ll gcd(ll a, ll b)
 9 {
10     return b == 0 ?  a :  gcd(b, a % b);
11 }
12 
13 int n, m;
14 int vis[N], used[N];
15 ll arr[N];
16 vector<int>vec;
17 
18 int main()
19 {
20     int t;
21     scanf("%d", &t);
22     for(int cas = 1; cas <= t; ++cas)
23     {
24         scanf("%d %d", &n, &m);
25         memset(vis, 0, sizeof vis);
26         memset(used, 0, sizeof used);
27         vec.clear();
28         for(int i = 1; i * i <= m; ++i)
29         {
30             if(m % i == 0)
31             {
32                 vec.push_back(i);
33                 if(i * i != m) vec.push_back(m / i);
34             }
35         }
36         sort(vec.begin(), vec.end());
37         int tot = vec.size();
38         tot--;
39         for(int i = 1; i <= n; ++i)
40         {
41             scanf("%lld", arr + i);
42             ll x = gcd(arr[i], m);
43             for(int j = 0; j < tot; ++j)
44             {
45                 if(vec[j] % x == 0) vis[j] = 1;
46             }
47         }
48         ll ans = 0;
49         for(int i = 0; i < tot; ++i)
50         {
51             if(vis[i] != used[i])
52             {
53                 ll tmp = (m - 1) / vec[i];
54                 ans += tmp * (tmp + 1) / 2 * vec[i] * (vis[i] - used[i]);
55                 for(int j = i + 1; j < tot; ++j)
56                 {
57                     if(vec[j] % vec[i] == 0) 
58                     {
59                         used[j] += vis[i] - used[i];
60                     }                        
61                 }
62             }
63         }
64         printf("Case #%d: %lld\n", cas, ans);
65     }
66     return 0;
67 }

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G - Game of Flying Circus

題意：翻譯啊翻譯

思路：分類討論

1)在1-2相遇   即v1=v2  輸出Yes

2)在2-3相遇，利用二分計算出相遇點，算一下到4的時間，判斷

3)在3-4相遇，利用二分計算出相遇點，算一下到1的時間，判斷

4)必定No

 1 #include<bits/stdc++.h>
 2 
 3 using namespace std;
 4 
 5 const double eps = 1e-8;
 6 
 7 double T, v1, v2;
 8 
 9 int sgn(double x)
10 {
11     if(fabs(x) < eps) return 0;
12     else return x > 0 ? 1 : -1;
13 }
14 
15 int main()
16 {
17     int t;
18     scanf("%d", &t);
19     for(int cas = 1; cas <= t; ++cas)
20     {
21         scanf("%lf %lf %lf", &T, &v1, &v2);
22         printf("Case #%d: ", cas);
23         if(sgn(v1 - v2) == 0) puts("Yes");
24         else if(sgn(2 * v1 * v1 - v2 * v2) > 0)//2-3
25         {
26             double l = 0, r = 300.0;
27             while(r - l > eps)
28             {
29                 double mid = (l + r) / 2.0;
30                 double dis1 = sqrt(mid * mid + 300.0 * 300.0);
31                 double t1 = dis1 / v1;
32                 double dis2 = mid + 300.0;
33                 double t2 = dis2 / v2;
34                 if(sgn(t1 - t2) > 0)
35                 {
36                     l = mid;
37                 }
38                 else 
39                 {
40                     r = mid;
41                 }
42             }
43             double x = (l + r) / 2.0;
44             double dis1 = x + 600.0;
45             double t1 = dis1 / v1;
46             double dis2 = 600.0 - x;
47             double t2 = dis2 / v2 + T;
48             puts(sgn(t1 - t2) <= 0 ? "Yes" : "No");
49         }
50         else if(sgn(3 * v1 - v2) > 0)//3-4
51         {
52             double l = 0, r = 300.0;
53             while(r - l > eps)
54             {
55                 double mid = (l + r) / 2.0;
56                 double dis1 = sqrt(mid * mid + 300.0 * 300.0);
57                 double t1 = dis1 / v1;
58                 double dis2 = 900.0 - mid;
59                 double t2 = dis2 / v2;
60                 if(sgn(t1 - t2) > 0)
61                 {
62                     r = mid;
63                 }
64                 else 
65                 {
66                     l = mid;
67                 }
68             }
69             double y = (l + r) / 2.0;
70             double dis1 = sqrt((300.0 - y) * (300.0 - y) + 300.0 * 300.0) + 900.0;
71             double t1 = dis1 / v1;
72             double dis2 = y + 300.0;
73             double t2 = dis2 / v2 + T;
74             puts(sgn(t1 - t2) <= 0 ? "Yes" : "No");
75         }
76         else puts("No");
77     }
78     return 0;
79 }

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H - Chessboard

留坑。

 

I - Triple

題意：二元組$A<a, b>$ 三元組 $B<c, d, e> A * B = {<a, c, d> | <a, b> \in A , <c, d, e> \in B and b = e}$ 求有多少集合知足TOP(C)

思路：枚舉B，對於每一個B，咱們考慮符合條件的最高的a, 由於有多個a，那麼小的a確定不會放到集合裏面，

對於最大的a，咱們判斷二維BIT中右下角有的最大值，是否大於它自己，不是的話它就應該放進TOP(C) 

要多考慮一下當前的(c, d) 有多個點的狀況

  1 #include <bits/stdc++.h>
  2 using namespace std;
  3 
  4 #define N 100010
  5 #define pii pair <int, int>
  6 #define ll long long
  7 
  8 int t, n, m;
  9 struct A
 10 {
 11     int a, b;
 12     void scan()
 13     {
 14         scanf("%d%d", &a, &b);
 15     }
 16     bool operator < (const A &r) const
 17     {
 18         return b == r.b ? a < r.a : b < r.b;
 19     }
 20 }a[N];
 21 
 22 struct B
 23 {
 24     int c, d, e;
 25     void scan()
 26     {
 27         scanf("%d%d%d", &c, &d, &e);
 28     }
 29 }b[N];
 30 
 31 pii pos[N];  
 32 
 33 struct BIT
 34 {
 35     int a[1010][1010];
 36 
 37     void Init()
 38     {
 39         memset(a, 0, sizeof a);
 40     }
 41 
 42     void update(int x, int y, int val)
 43     {
 44         for (int i = x; i; i -= i & -i)
 45             for (int j = y; j; j -= j & -j)
 46                 a[i][j] = max(a[i][j], val);
 47     }
 48 
 49     int query(int x, int y)
 50     {
 51         int res = 0;
 52         for (int i = x; i < 1010; i += i & -i)
 53             for (int j = y; j < 1010; j += j & -j)
 54                 res = max(res, a[i][j]);
 55         return res;
 56     }
 57 }bit; 
 58 
 59 int main()
 60 {
 61     scanf("%d", &t);
 62     for (int kase = 1; kase <= t; ++kase)
 63     {
 64         printf("Case #%d: ", kase); 
 65         scanf("%d%d", &n, &m); 
 66         for (int i = 1; i <= n; ++i) a[i].scan(); 
 67         for (int i = 1; i <= m; ++i) b[i].scan(); 
 68         sort(a + 1, a + 1 + n);     
 69         memset(pos, 0, sizeof pos); 
 70         for (int i = n; i >= 1; --i)
 71         {
 72             int b = a[i].b;
 73             if (!pos[b].first)
 74             {
 75                 pos[b].first = a[i].a, ++pos[b].second; 
 76                 continue;    
 77             }
 78             if (i != n) 
 79             {
 80                 if (b == a[i + 1].b && a[i].a == pos[b].first) 
 81                     ++pos[b].second; 
 82             }
 83         }
 84         bit.Init(); 
 85         for (int i = 1; i <= m; ++i) 
 86         {
 87             if (pos[b[i].e].first == 0) continue;
 88             bit.update(b[i].c, b[i].d, pos[b[i].e].first);
 89         }
 90         ll res = 0; 
 91         for (int i = 1; i <= m; ++i)
 92         {
 93             int x = b[i].c, y = b[i].d;
 94             if(!pos[b[i].e].first) continue;
 95             int Max = max(bit.query(x, y + 1), bit.query(x + 1, y));
 96             bool flag = true; 
 97             if (Max >= pos[b[i].e].first) flag = false;
 98             Max = bit.query(x, y); if (Max != pos[b[i].e].first) flag = false;
 99             if (flag) res += pos[b[i].e].second;
100         }
101         printf("%lld\n", res);
102     }
103     return 0;
104 }

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J - John's Fences

留坑。

 

K - Kykneion asma

留坑。

 

L - Number Link

留坑。

 

M - Meeting

題意：有多個集合，集合裏兩個點的距離都是$t-i$ 1 和 n  要到一個地方匯合，使得最大距離最小，求匯合點

思路：考慮暴力連邊的邊數最大有${\sum_{i = 1}^{i = m} S_i} ^2$ 無法解決

考慮將一個集合裏全部點向另外一個點連權值爲$t_i$的邊，再從這個點向集合裏全部點連權值爲0的邊

再對1和n分別跑最短路，枚舉中間點，更新答案

  1 #include <bits/stdc++.h>
  2 using namespace std;
  3 
  4 #define N 2000010
  5 #define ll long long
  6 #define INFLL 0x3f3f3f3f3f3f3f3f
  7 
  8 int t, n, m;
  9 int arr[N];
 10 
 11 struct Edge
 12 {
 13     int to, nx; ll w;
 14     Edge() {}
 15     Edge(int to, int nx, ll w) : to(to), nx(nx), w(w) {}
 16 }edge[N << 1]; 
 17 
 18 int head[N], pos;
 19 
 20 void Init()
 21 {
 22     memset(head, -1, sizeof head);
 23     pos = 0;
 24 }
 25 
 26 void addedge(int u, int v, int w)
 27 {
 28     edge[++pos] = Edge(v, head[u], w); head[u] = pos;
 29 }
 30 
 31 struct node
 32 {
 33     int u; ll w;
 34     node () {}
 35     node (int u, ll w) : u(u), w(w) {}
 36     bool operator < (const node &r) const
 37     {
 38         return w > r.w;
 39     }
 40 };
 41 
 42 ll dist[2][N];
 43 bool used[N];
 44 
 45 void Dijkstra(int st, int id)
 46 {
 47     for (int i = 1; i <= n + m; ++i) dist[id][i] = INFLL, used[i] = 0;
 48     priority_queue <node> q;
 49     q.emplace(st, 0);
 50     dist[id][st] = 0;
 51     while (!q.empty())
 52     {
 53         int u = q.top().u; q.pop();
 54         if (used[u]) continue;
 55         used[u] = 1;
 56         for (int it = head[u]; ~it; it = edge[it].nx)
 57         {
 58             int v = edge[it].to;
 59             if (!used[v] && dist[id][v] > dist[id][u] + edge[it].w)
 60             {
 61                 dist[id][v] = dist[id][u] + edge[it].w;
 62                 q.emplace(v, dist[id][v]);
 63             }
 64         }
 65     }
 66 }
 67 
 68 
 69 int main()
 70 {
 71     scanf("%d", &t);
 72     for (int kase = 1; kase <= t; ++kase)
 73     {
 74         printf("Case #%d: ", kase);
 75         scanf("%d%d", &n, &m);
 76         Init();
 77         for (int i = 1, tot, w; i <= m; ++i)
 78         {
 79             scanf("%d%d", &w, &tot);
 80             for (int j = 1; j <= tot; ++j) scanf("%d", arr + j);
 81             for (int j = 1; j <= tot; ++j) addedge(arr[j], i + n, w);
 82             for (int j = 1; j <= tot; ++j) addedge(i + n, arr[j], 0); 
 83         }    
 84         Dijkstra(1, 0);
 85         Dijkstra(n, 1);
 86         vector <int> v;
 87         ll Min = INFLL;
 88 //        for (int i = 1; i <= n; ++i) cout << dist[0][i] << " " << dist[1][i] << endl;
 89         for (int i = 1; i <= n; ++i) Min = min(Min, max(dist[0][i], dist[1][i]));
 90         if (Min == INFLL) puts("Evil John");
 91         else
 92         {
 93             for (int i = 1; i <= n; ++i) if (max(dist[0][i], dist[1][i]) == Min)
 94                 v.push_back(i);
 95             printf("%lld\n", Min);
 96             for (int i = 0, len = v.size(); i < len; ++i) printf("%d%c", v[i], " \n"[i == len - 1]);
 97         }
 98     }
 99     return 0;
100 }

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C=A∗B={⟨a,c,d⟩∣⟨a,b⟩∈A, ⟨c,d,e⟩∈B and b=e}